Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 5 \log_2 x = \log_2 y^3 - \log_{\sqrt{2}} 2 \\ \log_2 y = 8 - \log_{\sqrt{2}} x \end{cases}$

Given system of equations:

$5 \log_2 x = \log_2 y^3 - \log_{\sqrt{2}} 2$ $\;\;\; \cdots \; (1)$

$\log_2 y = 8 - \log_{\sqrt{2}} x$ $\;\;\; \cdots \; (2)$

We have from equation $(1)$, $\;$ $5 \log_2 x = 3 \log_2 y - \log_{\sqrt{2}} \left(\sqrt{2}\right)^2$

i.e. $\;$ $5 \log_2 x - 3 \log_2 y = - 2$ $\;\;\; \cdots \; (3)$

We have from equation $(2)$, $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\log_2 \sqrt{2}}$

i.e. $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\log_2 2^{\frac{1}{2}}}$

i.e. $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\dfrac{1}{2}}$

i.e. $\;$ $2 \log_2 x + \log_2 y = 8$ $\;\;\; \cdots \; (4)$

Multiplying equation $(4)$ by $3$ gives

$6 \log_2 x + 3 \log_2 y = 24$ $\;\;\; \cdots \; (5)$

Adding equations $(3)$ and $(5)$ gives

$11 \log_2 x = 22$

i.e. $\;$ $\log_2 x = 2$

i.e. $\;$ $x = 2^2 = 4$

Substituting the value of $x$ in equation $(4)$ gives

$2 \log_2 4 + \log_2 y = 8$

i.e. $\;$ $\log_2 y = 8 - 2 \log_2 2^2$

i.e. $\;$ $\log_2 y = 8 - 2 \times 2 = 4$

i.e. $\;$ $y = 2^4 = 16$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(4, \; 16\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} x + y = 4 + \sqrt{y^2 + 2} \\ \log x - 2 \log 2 = \log \left(1 + \dfrac{1}{2} y\right) \end{cases}$

Given system of equations:

$x + y = 4 + \sqrt{y^2 + 2}$ $\;\;\; \cdots \; (1)$

$\log x - 2 \log 2 = \log \left(1 + \dfrac{1}{2} y\right)$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$,

$\log x - \log 4 = \log \left(1 + \dfrac{y}{2}\right)$

i.e. $\;$ $\log \left(\dfrac{x}{4}\right) = \log \left(1 + \dfrac{y}{2}\right)$ $\;\;\; \cdots \; (3)$

Taking antilog on both sides of equation $(3)$ gives

$\dfrac{x}{4} = 1 + \dfrac{y}{2}$

i.e. $\;$ $x - 4 = 2y$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$ equation $(1)$ becomes

$2y + y = \sqrt{y^2 +2}$

i.e. $\;$ $3y = \sqrt{y^2 + 2}$

i.e. $\;$ $9y^2 = y^2 + 2$

i.e. $\;$ $8y^2 = 2$

i.e. $\;$ $y^2 = \dfrac{1}{4}$ $\implies$ $y = \pm \dfrac{1}{2}$

When $y = \dfrac{-1}{2}$, we have from equation $(4)$

$x = 4 - 2 \times \dfrac{1}{2} = 3$

Substituting $\left(x, y\right) = \left(3, \dfrac{-1}{2}\right)$ in equation $(1)$ gives

$3 - \dfrac{1}{2} = 4 + \sqrt{\dfrac{1}{4} + 2}$

i.e. $\;$ $\dfrac{5}{2} = 4 + \dfrac{3}{2}$

i.e. $\;$ $\dfrac{5}{2} = \dfrac{11}{2}$ $\;$ which is not true.

$\implies$ $\left(3, \dfrac{-1}{2}\right)$ is not a valid solution.

When $y = \dfrac{1}{2}$, we have from equation $(4)$

$x = 4 + 2 \times \dfrac{1}{2} = 5$

Substituting $\left(x, y\right) = \left(5, \dfrac{1}{2}\right)$ in equation $(1)$ gives

$5 + \dfrac{1}{2} = 4 + \sqrt{\dfrac{1}{4} + 2}$

i.e. $\;$ $\dfrac{11}{2} = 4 + \dfrac{3}{2}$

i.e. $\;$ $\dfrac{11}{2} = \dfrac{11}{2}$ $\;$ which is true.

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(5, \; \dfrac{1}{2}\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \\ \log_3 x + \log_3 y = 1 + \log_3 5 \end{cases}$

Given system of equations:

$\dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15}$ $\;\;\; \cdots \; (1)$

$\log_3 x + \log_3 y = 1 + \log_3 5$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$,

$\log_3 x + \log_3 y - \log_3 5 = 1$

i.e. $\;$ $\log_3 \left(\dfrac{x \; y}{5}\right) = 1$

i.e. $\;$ $\dfrac{x \; y}{5} = 3^1 = 3$

$\implies$ $x = \dfrac{15}{y}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$\dfrac{y}{15} - \dfrac{1}{y} = \dfrac{2}{15}$

i.e. $\;$ $y^2 - 2y - 15 = 0$

i.e. $\;$ $\left(y - 5\right) \left(y + 3\right) = 0$

i.e. $\;$ $y = 5$ $\;$ or $\;$ $y = -3$

When $y = -3$, the term $\log_3 y$ in equation $(2)$ becomes $\log_3 \left(-3\right)$.

But logarithm of a negative number is not defined.

$\therefore \;$ $y = -3$ is not a valid solution.

When $y = 5$, we have from equation $(3)$, $\;\;$ $x = \dfrac{15}{5} = 3$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(3, \; 5\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} \log_2 x + 2 \log_2 y = 3 \\ x^2 + y^4 = 16 \end{cases}$

Given system of equations:

$\log_2 x + 2 \log_2 y = 3$ $\;\;\; \cdots \; (1)$

$x^2 + y^4 = 16$ $\;\;\; \cdots \; (2)$

Equation $(1)$ $\implies$ $\log_2 x + \log_2 y^2 = 3$

i.e. $\;$ $\log_2 \left(x y^2\right) = 3$

i.e. $\;$ $x y^2 = 2^3 = 8$

i.e. $\;$ $y^2 = \dfrac{8}{x}$ $\implies$ $y^4 = \dfrac{64}{x^2}$ $\;\;\; \cdots \; (3)$

Substituting the value of $y^4$ from equation $(3)$ in equation $(2)$ gives

$x^2 + \dfrac{64}{x^2} = 16$

i.e. $\;$ $x^4 - 16 x^2 + 64 = 0$

i.e. $\;$ $\left(x^2 - 8\right)^2 = 0$

i.e. $\;$ $x^2 - 8 = 0$

i.e. $\;$ $x^2 = 8$ $\implies$ $x = \pm 2 \sqrt{2}$

When $x = -2 \sqrt{2}$, the term $\log_2 x$ in equation $(1)$ becomes $\log_2 \left(-2\sqrt{2}\right)$.

But logarithm of a negative number is not defined.

$\implies$ $x - 2 \sqrt{2}$ is not a valid solution.

When $x = + 2 \sqrt{2}$, we have from from equation $(1)$

$\log_2 2 \sqrt{2} + 2 \log_2 y = 3$

i.e. $\;$ $\log_2 2 + \log_2 2^{\frac{1}{2}} + 2 \log_2 y = 3$

i.e. $\;$ $1 + \dfrac{1}{2} + 2 \log_2 y = 3$

i.e. $\;$ $2 \log_2 y = \dfrac{3}{2}$

i.e. $\;$ $\log_2 y = \dfrac{3}{4}$

$\implies$ $y = 2^{\frac{3}{4}} = \left(2^3\right)^{\frac{1}{4}} = \sqrt[4]{8}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2 \sqrt{2}, \; \sqrt[4]{8}\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} \log_4 x - \log_x y = \dfrac{7}{6} \\ xy = 16 \end{cases}$

Given system of equations:

$\log_4 x - \log_x y = \dfrac{7}{6}$ $\;\;\; \cdots \; (1)$

$xy = 16$ $\;\;\; \cdots \; (2)$

Equation $(1)$ $\implies$ $6 \log_4 x - 6 \log_x y = 7$ $\;\;\; \cdots \; (3)$

Equation $(2)$ $\implies$ $y = \dfrac{16}{x}$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(3)$ becomes

$6 \log_4 x - 6 \log_x \left(\dfrac{16}{x}\right) = 7$

i.e. $\;$ $6 \log_4 x - 6 \log_x 16 + 6 \log_x x = 7$

i.e. $\;$ $6 \log_4 x - 6 \left(\dfrac{\log_4 16}{\log_4 x}\right) + 6 \times 1 = 7$

i.e. $\;$ $6 \log_4 x - 6 \left(\dfrac{2}{\log_4 x}\right) = 1$

i.e. $\;$ $6 \left(\log_4 x\right)^2 - \log_4 x - 12 = 0$

i.e. $\;$ $6 \left(\log_4 x\right)^2 - 9 \log_4 x + 8 \log_4 x - 12 = 0$

i.e. $\;$ $3 \log_4 x \left(2 \log_4 x - 3\right) + 4 \left(2 \log_4 x - 3\right) = 0$

i.e. $\;$ $\left(3 \log_4 x + 4\right) \left(2 \log_4 x - 3\right) = 0$

i.e. $\;$ $\log_4 x = \dfrac{-4}{3}$ $\;$ or $\;$ $\log_4 x = \dfrac{3}{2}$

i.e. $\;$ $x = 4^{\frac{-4}{3}} = \left(4^{-4}\right)^{\frac{1}{3}} = \dfrac{1}{\sqrt[3]{256}}$ $\;$ is an irrational number

or $\;$ $x = 4^{\frac{3}{2}} = \left(4^3\right)^{\frac{1}{2}} = \sqrt{64} = 8$

When $\;$ $x = \dfrac{1}{\sqrt[3]{256}}$, $\;$ the base of the term $\;$ $\log_x y$ $\;$ in equation $(1)$ becomes irrational.

But, base of a logarithm cannot be irrational.

$\therefore \;$ $x = \dfrac{1}{\sqrt[3]{256}}$ $\;$ is not a valid solution.

When $\;$ $x = 8$, $\;$ we have from equation $(4)$, $\;$ $y = \dfrac{16}{8} = 2$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(8, \; 2\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} x^{\log y} = 2 \\ xy = 20 \end{cases}$

Given system of equations:

$x^{\log y} = 2$ $\;\;\; \cdots \; (1)$

$xy = 20$ $\;\;\; \cdots \; (2)$

Taking log on both sides of equation $(1)$ gives

$\log \left(x^{\log y}\right) = \log 2$

i.e. $\;$ $\log y \times \log x = \log 2$ $\;\;\; \cdots \; (3)$

Equation $(2)$ $\implies$ $x = \dfrac{20}{y}$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(3)$ becomes

$\log y \times \log \left(\dfrac{20}{y}\right) = \log 2$

i.e. $\;$ $\log y \left[\log 20 - \log y\right] = \log 2$

i.e. $\;$ $\left(\log y\right)^2 - \log 20 \; \log y + \log 2 = 0$

i.e. $\;$ $\left(\log y\right)^2 - \left[\log \left(10 \times 2\right)\right] \log y + \log 2 = 0$

i.e. $\;$ $\left(\log y\right)^2 - \left[\log 10 + \log 2\right] \log y + \log 2 = 0$

i.e. $\;$ $\left(\log y\right)^2 - \left[1 + \log 2\right] \log y + \log 2 = 0$

i.e. $\;$ $\left(\log y\right)^2 - \log y - \log 2 \; \log y + \log 2 = 0$

i.e. $\;$ $\log y \left(\log y - 1\right) - \log 2 \left(\log y - 1\right) = 0$

i.e. $\;$ $\left(\log y -1\right) \left(\log y - \log 2\right) = 0$

i.e. $\;$ $\log y = 1$ $\;$ or $\;$ $\log y = \log 2$

i.e. $\;$ $y = 10^1 = 10$ $\;$ or $\;$ $y = 2$

When $\;$ $y = 10$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{20}{10} = 2$

When $\;$ $y = 2$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{20}{2} = 10$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, \; 10\right), \; \left(10, \; 2\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} y - \log_3 x = 1 \\ x^y = 3^{12} \end{cases}$

Given system of equations:

$y - \log_3 x = 1$ $\;\;\; \cdots \; (1)$

$x^y = 3^{12}$ $\;\;\; \cdots \; (2)$

Equation $(1)$ $\implies$ $\log_3 x = y - 1$

i.e. $\;$ $x = 3^{y - 1}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(2)$ becomes

$\left(3^{y - 1}\right)^y = 3^{12}$

i.e. $\;$ $y \left(y - 1\right) = 12$

i.e. $\;$ $y^2 - y - 12 = 0$

i.e. $\;$ $\left(y - 4\right) \left(y + 3\right) = 0$

i.e. $\;$ $y = 4$ $\;$ or $\;$ $y = -3$

When $\;$ $y = 4$, $\;$ we have from equation $(3)$

$x = 3^{4 - 1} = 3^3 = 27$

When $\;$ $y = -3$, $\;$ we have from equation $(3)$

$x = 3^{-3 - 1} = 3^{-4} = \dfrac{1}{81}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(27, \; 4\right), \; \left(\dfrac{1}{81}, \; -3\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 3^x \times 2^y = 972 \\ \log_{\sqrt{3}} \left(x - y\right) = 2 \end{cases}$

Given system of equations:

$3^x \times 2^y = 972$ $\;\;\; \cdots \; (1)$

$\log_{\sqrt{3}} \left(x - y\right) = 2$ $\;\;\; \cdots \; (2)$

Equation $(2)$ $\implies$ $x - y = \sqrt{3}^2 = 3$

i.e. $\;$ $x = y + 3$ $\;\;\; \cdots \; (3)$

Substituting the value of $x$ from equation $(3)$ in equation $(1)$ gives

$3^{y + 3} \times 2^y = 972$

i.e. $\;$ $3^y \times 3^3 \times 2^y = 972$

i.e. $\;$ $6^y = \dfrac{972}{27} = 36 = 6^2$

$\implies$ $y = 2$

Substituting the value of $y$ in equation $(3)$ gives $\;\;$ $x = 2 + 3 = 5$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(5, \; 2\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 4^{\frac{x}{y}} - 3 \times 4^{\frac{5y-x}{y}} = 16 \\ \sqrt{x} - \sqrt{2y} = \sqrt{12} - \sqrt{8} \end{cases}$

Given system of equations:

$4^{\frac{x}{y}} - 3 \times 4^{\frac{5y-x}{y}} = 16$ $\;\;\; \cdots \; (1)$

$\sqrt{x} - \sqrt{2y} = \sqrt{12} - \sqrt{8}$ $\;\;\; \cdots \; (2)$

We have from equation $(1)$

$4^{\frac{x}{y}} - 3 \times 4^5 \times 4^{\frac{-x}{y}} = 16$

i.e. $\;$ $4^{\frac{x}{y}} - \dfrac{3072}{4^{\frac{x}{y}}} = 16$

i.e. $\;$ $\left(4^{\frac{x}{y}}\right)^2 - 16 \times 4^{\frac{x}{y}} - 3072 = 0$

i.e. $\;$ $\left(4^{\frac{x}{y}}\right)^2 - 64 \times 4^{\frac{x}{y}} + 48 \times 4^{\frac{x}{y}} - 3072 = 0$

i.e. $\;$ $4^{\frac{x}{y}} \left(4^{\frac{x}{y}} - 64\right) + 48 \left(4^{\frac{x}{y}} - 64\right) = 0$

i.e. $\;$ $\left(4^{\frac{x}{y}} - 64\right) \left(4^{\frac{x}{y}} + 48\right) = 0$

i.e. $\;$ $4^{\frac{x}{y}} = 64$ $\;$ or $\;$ $4^{\frac{x}{y}} = -48$

i.e. $\;$ $\dfrac{x}{y} = \log_4 64$ $\;$ or $\;$ $\dfrac{x}{y} = \log_4 \left(-48\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $\dfrac{x}{y} = \log_4 \left(-48\right)$ $\;$ is not a valid solution.

Now, $\;$ $\dfrac{x}{y} = \log_4 64$ $\implies$ $\dfrac{x}{y} = \log_4 4^3$

i.e. $\;$ $\dfrac{x}{y} = 3 \log_4 4 = 3$

$\implies$ $x = 3y$ $\;\;\; \cdots \; (3)$

Substituting the value of $x$ in equation $(2)$ gives

$\sqrt{3y} - \sqrt{2y} = \sqrt{12} - \sqrt{8}$

i.e. $\;$ $\sqrt{y} \left(\sqrt{3} - \sqrt{2}\right) = 2 \sqrt{3} - 2 \sqrt{2}$

i.e. $\;$ $\sqrt{y} \left(\sqrt{3} - \sqrt{2}\right) = 2 \left(\sqrt{3} - \sqrt{2}\right)$

i.e. $\;$ $\sqrt{y} = 2$ $\implies$ $y = 4$

Substituting the value of $y$ in equation $(3)$ gives $\;$ $x = 3 \times 4 = 12$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(12, \; 4\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 7^x - 16y = 0 \\ 4^x - 49y = 0 \end{cases}$

Given system of equations:

$7^x - 16y = 0$ $\;\;\; \cdots \; (1)$

$4^x - 49y = 0$ $\;\;\; \cdots \; (2)$

Multiplying equation $(1)$ with $49$ gives

$49 \times 7^x - 784 y = 0$

i.e. $\;$ $7^2 \times7^x - 784 y = 0$

i.e. $\;$ $7^{x + 2} = 784 y$ $\;\;\; \cdots \; (1a)$

Multiplying equation $(2)$ with $16$ gives

$16 \times 4^x - 784 y = 0$

i.e. $\;$ $4^2 \times4^x - 784 y = 0$

i.e. $\;$ $4^{x + 2} = 784 y$ $\;\;\; \cdots \; (2a)$

We have from equations $(1)$ and $(2)$

$7^{x+2} = 4^{x+2}$ $\;\;\; \cdots \; (3)$

i.e. $\;$ $\dfrac{7^{x+2}}{4^{x+2}} = 1$

i.e. $\;$ $\left(\dfrac{7}{4}\right)^{x+2} = 1$

i.e. $\;$ $x + 2 = \log_{\frac{7}{4}} 1$

i.e. $\;$ $x + 2 = 0$ $\implies$ $x = - 2$

Substituting the value of $x$ in equation $(1a)$ gives

$7^{-2 + 2} = 784 y$

i.e. $\;$ $7^0 = 784 y$

i.e. $\;$ $784 y = 1$ $\implies$ $y = \dfrac{1}{784}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(-2, \; \dfrac{1}{784}\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 3 \times 2^x + 2 \times 3^y = 2.75 \\ 2^x - 3^y = -0.75 \end{cases}$

Given system of equations:

$3 \times 2^x + 2 \times 3^y = 2.75$ $\;\;\; \cdots \; (1)$

$2^x - 3^y = -0.75$ $\;\;\; \cdots \; (2)$

Multiply equation $(2)$ with $2$ and adding to equation $(1)$ gives

$5 \times 2^x = 1.25$

i.e. $\;$ $2^x = 0.25 = \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2 = 2^{-2}$

i.e. $\;$ $x = \log_2 2^{-2} = -2 \log_2 2 = -2$

Substituting the value of $x$ in equation $(2)$ gives

$2^{-2} - 3^y = -0.75$

i.e. $\;$ $3^y = 0.25 + 0.75 = 1$

i.e. $\;$ $y = \log_3 1 = 0$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(-2, 0\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} \log_4 x - \log_2 y = 0 \\ x^2 - 5y^2 + 4 = 0 \end{cases}$

Given system of equations:

$\log_4 x - \log_2 y = 0$ $\;\;\; \cdots \; (1)$

$x^2 - 5y^2 + 4 = 0$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as $\;\;$ $\dfrac{\log_2 x}{\log_2 4} = \log_2 y$

i.e. $\;$ $\dfrac{\log_2 x}{\log_2 2^2} = \log_2 y$

i.e. $\;$ $\dfrac{\log_2 x}{2 \log_2 2} = \log_2 y$

i.e. $\;$ $\dfrac{\log_2 x}{2} = \log_2 y$

i.e. $\;$ $\log_2 x = 2 \log_2 y$

i.e. $\;$ $\log_2 x = \log_2 y^2$ $\;\;\; \cdots \; (1a)$

Taking antilog to the base $2$ on both sides of equation $(1a)$ gives

$x = y^2$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(2)$ becomes

$x^2 - 5x^2 + 4 = 0$

i.e. $\;$ $\left(x - 4\right) \left(x - 1\right) = 0$

i.e. $\;$ $x = 4$ $\;$ or $\;$ $x = 1$

Substituting the values of $x$ in equation $(3)$ give

when $\;$ $x = 4$ $\implies$ $y^2 = 4$ $\implies$ $y = \pm 2$

when $\;$ $x = 1$ $\implies$ $y^2 = 1$ $\implies$ $y = \pm 1$

When $\;$ $y = -2$ $\;$ or $\;$ $y = -1$, $\;$ the term $\;$ $\log_2 y$ $\;$ in equation $(1)$ becomes $\;$ $\log_2 \left(-2\right)$ $\;$ and $\;$ $\log_2 \left(-1\right)$ $\;$ respectively.

But logarithm of a negative number is not defined.

$\implies$ $y = -2$ $\;$ and $\;$ $y = -1$ $\;$ are not valid solutions to the given system of equations.

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(4, 2\right), \; \left(1, 1\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 2^x + 2^y = 12 \\ x + y = 5 \end{cases}$

Given system of equations:

$2^x + 2^y = 12$ $\;\;\; \cdots \; (1)$

$x + y = 5$ $\;\;\; \cdots \; (2)$

Let $\;$ $2^x = p$ $\;\;\; \cdots \; (3a)$

$\implies$ $x = \log_2 p$ $\;\;\; \cdots \; (3b)$

and $\;$ $2^y = q$ $\;\;\; \cdots \; (4a)$

$\implies$ $y = \log_2 q$ $\;\;\; \cdots \; (4b)$

In view of equations $(3a)$ and $(4a)$ equation $(1)$ becomes

$p + q = 12$ $\;\;\; \cdots \; (5)$

In view of equations $(3b)$ and $(4b)$ equation $(2)$ becomes

$\log_2 p + \log_ q = 5$

i.e. $\;$ $\log_2 \left(pq\right) = 5$

i.e. $\;$ $pq = 2^5 = 32$ $\;\;\; \cdots \; (6)$

Now, $\;$ $\left(p - q\right)^2 = \left(p + q\right)^2 - 4pq$

i.e. $\;$ $\left(p - q\right)^2 = 144 - 128 = 16$ $\;\;\;$ [by equations $(5)$ and $(6)$]

i.e. $\;$ $p - q = 4$ $\;\;\; \cdots \; (7a)$ $\;$ or $\;$ $p - q = -4$ $\;\;\; \cdots \; (7b)$

Solving equations $(5)$ and $(7a)$ simultaneously gives

$2p = 16$ $\implies$ $p = 8$ $\;\;\; \cdots \; (8a)$

Substituting the value of $p$ from equation $(8a)$ in equation $(5)$ gives

$q = 4$ $\;\;\; \cdots \; (8b)$

Solving equations $(5)$ and $(7b)$ simultaneously gives

$2p = 8$ $\implies$ $p = 4$ $\;\;\; \cdots \; (9a)$

Substituting the value of $p$ from equation $(9a)$ in equation $(5)$ gives

$q = 8$ $\;\;\; \cdots \; (9b)$

In view of equations $(8a)$ and $(8b)$, equations $(3b)$ and $(4b)$ become

$x = \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3$

$y = \log_2 4 = \log_2 2^2 = 2 \log_2 2 = 2$

In view of equations $(9a)$ and $(9b)$, equations $(3b)$ and $(4b)$ become

$x = \log_2 4 = \log_2 2^2 = 2 \log_2 2 = 2$

$y = \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, 3\right), \; \left(3, 2\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 3^{2x} - 2^y = 77 \\ 3^x - 2^{\frac{y}{2}} = 7 \end{cases}$

Given system of equations:

$3^{2x} - 2^y = 77$ $\;\;\; \cdots \; (1)$

$3^x - 2^{\frac{y}{2}} = 7$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$\left(3^x\right)^2 - \left(2^{\frac{y}{2}}\right)^2 = 77$

i.e. $\;$ $\left(3^x + 2 ^{\frac{y}{2}}\right) \left(3^x - 2^{\frac{y}{2}}\right) = 77$

i.e. $\;$ $\left(3^x + 2^{\frac{y}{2}}\right) \times 7 = 77$ $\;\;\;$ [in view of equation $(2)$]

i.e. $\;$ $3^x + 2^{\frac{y}{2}} = 11$ $\;\;\; \cdots \; (3)$

Adding equations $(2)$ and $(3)$ gives

$2 \times 3^x = 18$

i.e. $\;$ $3^x = 9 = 3^2$

$\implies$ $x = 2$

Substituting the value of $x$ in equation $(3)$ gives

$3^2 + 2^{\frac{y}{2}} = 11$

i.e. $\;$ $2^{\frac{y}{2}} = 2 = 2^1$

$\implies$ $\dfrac{y}{2} = 1$ $\implies$ $y = 2$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, 2\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 4^{\left(x-y\right)^2 - 1} = 1 \\ 5^{x + y} = 125 \end{cases}$

Given system of equations:

$4^{\left(x-y\right)^2 - 1} = 1$ $\;\;\; \cdots \; (1)$

$5^{x + y} = 125$ $\;\;\; \cdots \; (2)$

Equation $(1)$ $\implies$ $\left(x - y\right)^2 - 1 = \log_4 1$

i.e. $\;$ $\left(x - y\right)^2 - 1 = 0$

i.e. $\;$ $\left(x - y\right)^2 = 1$

i.e. $\;$ $x - y = 1$ $\;\;\; \cdots \; (3a)$ $\;$ or $\;$ $x - y = -1$ $\;\;\; \cdots \; (3b)$

Equation $(2)$ $\implies$ $x + y = \log_5 125$

i.e. $\;$ $x + y = \log_5 5^3$

i.e. $\;$ $x + y = 3 \log_5 5$

i.e. $\;$ $x + y = 3$ $\;\;\; \cdots \; (4)$

Solving equations $(3a)$ and $(4)$ simultaneously [adding equations $(3a)$ and $(4)$] gives

$2x = 4$ $\implies$ $x = 2$

Substituting the value of $x$ in equation $(4)$ gives

$y = 3 - 2 = 1$

Solving equations $(3b)$ and $(4)$ simultaneously [adding equations $(3b)$ and $(4)$] gives

$2x = 2$ $\implies$ $x = 1$

Substituting the value of $x$ in equation $(4)$ gives

$y = 3 - 1 = 2$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(1, 2\right), \; \left(2, 1\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} 4^{x+y} = 128 \\ 5^{3x-2y-3} = 1 \end{cases}$

Given system of equations:

$4^{x+y} = 128$ $\;\;\; \cdots \; (1)$

$5^{3x-2y-3} = 1$ $\;\;\; \cdots \; (2)$

Equation $(1)$ $\implies$ $x + y = \log_4 128$

i.e. $\;$ $x + y = \log_4 2^7$

i.e. $\;$ $x + y = 7 \log_4 2$

i.e. $\;$ $x + y = 7 \log_4 4^{\frac{1}{2}}$

i.e. $\;$ $x + y = \dfrac{7}{2} \log_4 4$

i.e. $\;$ $x + y = \dfrac{7}{2}$

i.e. $\;$ $2x + 2y = 7$ $\;\;\; \cdots \; (3)$

Equation $(2)$ $\implies$ $3x - 2y - 3 = \log_5 1$

i.e. $\;$ $3x - 2y - 3 = 0$

i.e. $\;$ $3x - 2y = 3$ $\;\;\; \cdots \; (4)$

Solving equations $(3)$ and $(4)$ simultaneously [adding equations $(3)$ and $(4)$] gives

$5x = 10$ $\implies$ $x = 2$

Substituting the value of $x$ in equation $(3)$ gives

$y = \dfrac{7 - 4}{2} = \dfrac{3}{2}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, \dfrac{3}{2}\right) \right\}$

Algebra - Systems of Exponential and Logarithmic Equations

Solve the following system of equations: $\begin{cases} \log \sqrt{5 - x} + \log 2 = \log \left(x + 3\right) \\ x^2 + 7x - 8 = 0 \end{cases}$

Given system of equations:

$\log \sqrt{5 - x} + \log 2 = \log \left(x + 3\right)$ $\;\;\; \cdots \; (1)$

$x^2 + 7x - 8 = 0$ $\;\;\; \cdots \; (2)$

Solving equation $(2)$ gives

$\left(x + 8\right) \left(x - 1\right) = 0$

i.e. $\;$ $x = -8$ $\;$ or $\;$ $x = 1$

When $\;$ $x = -8$, $\;$ the term $\;$ $\log \left(x + 3\right)$ $\;$ in equation $(1)$ becomes $\;$ $\log \left(-8 + 3\right) = \log \left(-5\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $x = -8$ $\;$ is not a valid solution to the given pair of equations.

Substituting $\;$ $x = 1$ $\;$ in equation $(1)$ gives

$\log \sqrt{5 - 1} + \log 2 = \log \left(1 + 3\right)$

i.e. $\;$ $\log \sqrt{4} + \log 2 = \log 4$

i.e. $\;$ $\log 2 + \log 2 = \log 4$

i.e. $\;$ $2 \log 2 = \log 4$

i.e. $\;$ $\log 2^2 = \log 4$ $\implies$ $\log 4 = \log 4$

$\therefore \;$ $x = 1$ $\;$ satisfies both equations $(1)$ and $(2)$.

$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{1 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $2 \log_2 \left(\dfrac{x - 7}{x - 1}\right) + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$

Given equation: $\;\;$ $2 \log_2 \left(\dfrac{x - 7}{x - 1}\right) + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$

i.e. $\;$ $\log_2 \left(\dfrac{x - 7}{x - 1}\right)^2 + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$

i.e. $\;$ $\log_2 \left[\left(\dfrac{x - 7}{x - 1}\right)^2 \times \left(\dfrac{x - 1}{x + 1}\right)\right] = 1$

i.e. $\;$ $\dfrac{\left(x - 7\right)^2}{\left(x + 1\right) \left(x - 1\right)} = 2^1 = 2$

i.e. $\;$ $x^2 - 14x + 49 = 2x^2 - 2$

i.e. $\;$ $x^2 + 14x - 51 = 0$

i.e. $\;$ $\left(x + 17\right) \left(x - 3\right) = 0$

i.e. $\;$ $x + 17 = 0$ $\;$ or $\;$ $x - 3 = 0$

i.e. $\;$ $x = -17$ $\;$ or $\;$ $x = 3$

When $\;$ $x = 3$, $\;$ the term $\;$ $\log_2 \left(\dfrac{x - 7}{x - 1}\right)$ $\;$ in the given problem becomes

$\log_2 \left(\dfrac{3 - 7}{3 - 1}\right) = \log \left(-2\right)$

But, logarithm of a negative number is not defined.

When $\;$ $x = -17$, $\;$ the given problem becomes

$2 \log_2 \left(\dfrac{-17 - 7}{-17 - 1}\right) + \log_2 \left(\dfrac{-17 - 1}{-17 + 1}\right) = 1$

i.e. $\;$ $\log_2 \left(\dfrac{-24}{-18}\right)^2 + \log_2 \left(\dfrac{-18}{-16}\right) = 1$

i.e. $\;$ $\log_2 \left(\dfrac{4}{3}\right)^2 + \log_2 \left(\dfrac{9}{8}\right) = 1$

i.e. $\;$ $\log_2 \left(\dfrac{16}{9} \times \dfrac{9}{8}\right) = 1$

i.e. $\;$ $\log_2 2 = 1$

i.e. $\;$ $1 = 1$ $\;\;$ which is true.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-17 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $5^{\log x} = 50 - x^{\log 5}$

Given equation: $\;\;$ $5^{\log x} = 50 - x^{\log 5}$ $\;\;\; \cdots \; (1)$

Let $\;$ $5^{\log x} = p$

Then, by definition $\;$ $\log_5 p = \log x$ $\;\;\; \cdots \; (2)$

$\implies$ $x = 10^{\log_5 p}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$p = 50 - \left(10^{\log_5 p}\right)^{\log 5}$

i.e. $\;$ $p = 50 - \left[10^{\left(\frac{\log p}{\log 5}\right)}\right]^{\log 5}$

i.e. $\;$ $p = 50 - 10^{\log p}$

i.e. $\;$ $p = 50 - p$

i.e. $\;$ $2p = 50$ $\implies$ $p = 25$

Substituting the value of $p$ in equation $(2)$ gives

$\log_5 25 = \log x$

i.e. $\;$ $\log x = \log_5 5^2 = 2 \log_5 5 = 2$

$\implies$ $x = 10^2 = 100$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{100 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log^2 \left(100x\right) - \log^2 \left(10 x\right) + \log^2 x = 6$

Given equation: $\;\;$ $\log^2 \left(100x\right) - \log^2 \left(10 x\right) + \log^2 x = 6$

i.e. $\;$ $\left(\log 100 + \log x\right)^2 - \left(\log 10 + \log x\right)^2 + \log^2 x = 6$

i.e. $\;$ $\left(2 + \log x\right)^2 - \left(1 + \log x\right)^2 + \log^2 x = 6$

i.e. $\;$ $\log^2 x + 4 \log x + 4 - \left(\log^2 x + 2 \log x + 1\right) + \log^2 x = 6$

i.e. $\;$ $\log^2 x + 2 \log x - 3 = 0$

i.e. $\;$ $\log^2 x + 3 \log x - \log x - 3 = 0$

i.e. $\;$ $\log x \left(\log x + 3\right) - 1 \left(\log x + 3\right) = 0$

i.e. $\;$ $\left(\log x + 3\right) \left(\log x - 1\right) = 0$

i.e. $\;$ $\log x + 3 = 0$ $\;$ or $\;$ $\log x - 1 = 0$

i.e. $\;$ $\log x = -3$ $\;$ or $\;$ $\log x = 1$

i.e. $\;$ $x = 10^{-3}$ $\;$ or $\;$ $x = 10^1 = 10$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-3}, \; 10 \right\}$