Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(3x^2 + 7\right) - \log \left(3x - 2\right) = 1$

Given equation: $\;\;$ $\log \left(3x^2 + 7\right) - \log \left(3x - 2\right) = 1$

i.e. $\;$ $\log \left(\dfrac{3x^2 + 7}{3x - 2}\right) = 1$

i.e. $\;$ $\dfrac{3x^2 + 7}{3x - 2} = 10^1 = 10$

i.e. $\;$ $3x^2 + 7 = 30x - 20$

i.e. $\;$ $3x^2 - 30x + 27 = 0$

i.e. $\;$ $x^2 - 10x + 9 = 0$

i.e. $\;$ $\left(x - 9\right) \left(x - 1\right) = 0$

i.e. $\;$ $x = 9$, $\;$ or $\;$ $x = 1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1, 9 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{1}{2} \log x + 3 \log \sqrt{2 + x} = \log \sqrt{x \left(x + 2\right)} + 2$

Given equation: $\;\;$ $\dfrac{1}{2} \log x + 3 \log \sqrt{2 + x} = \log \sqrt{x \left(x + 2\right)} + 2$

i.e. $\;$ $\log \sqrt{x} - \log \sqrt{x \left(x + 2\right)} + \log \left(\sqrt{2 + x}\right)^3 = 2$

i.e. $\;$ $\log \left[\dfrac{\sqrt{x} \left(2 + x\right) \sqrt{2 + x}}{\sqrt{x \left(x + 2\right)}}\right] = 2$

i.e. $\;$ $\log \left(x + 2\right) = 2$

i.e. $\;$ $x +2 = 10^2 = 100$

i.e. $\;$ $x = 98$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{98 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{\log \sqrt{x + 7} - \log 2}{\log 8 - \log \left(x - 5\right)} = -1$

Given equation: $\;\;$ $\dfrac{\log \sqrt{x + 7} - \log 2}{\log 8 - \log \left(x - 5\right)} = -1$

i.e. $\;$ $\log \sqrt{x+7} - \log 2 = \log \left(x-5\right) - \log 8$

i.e. $\;$ $\log \sqrt{x+7} - \log \left(x - 5\right) - \log 2 + 3 \log 2 = 0$

i.e. $\;$ $\log \sqrt{x+7} - \log \left(x-5\right) + 2 \log 2 = 0$

i.e. $\;$ $\log \sqrt{x + 7} - \log \left(x - 5\right) + \log 4 = 0$

i.e. $\;$ $\log \left(\dfrac{4 \sqrt{x + 7}}{x - 5}\right) = 0$

i.e. $\;$ $\dfrac{4 \sqrt{x + 7}}{x - 5} = 10^0 = 1$

i.e. $\;$ $4 \sqrt{x + 7} = x - 5$

i.e. $\;$ $16 \left(x + 7\right) = x^2 - 10 x + 25$

i.e. $\;$ $x^2 - 26 x - 87 = 0$

i.e. $\;$ $\left(x - 29\right) \left(x + 3\right) = 0$

i.e. $\;$ $x = 29$ $\;$ or $\;$ $x = -3$

But, when $\;$ $x = -3$, $\;$ the term $\;$ $\log \left(x - 5\right)$ $\;$ in the given problem becomes

$\log \left(-3 - 5\right) = \log \left(-8\right)$

$\because \;$ logarithm of a negative number is not defined,

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

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{29 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(4^x + 1\right) = x + \log_2 \left(2^{x+3} - 6\right)$

Given equation: $\;\;$ $\log_2 \left(4^x + 1\right) = x + \log_2 \left(2^{x+3} - 6\right)$

i.e. $\;$ $\log_2 \left(4^x + 1\right) - \log_2 \left(2^{x+3} - 6\right) = x$

i.e. $\;$ $\log_2 \left[\dfrac{4^x + 1}{2^{x+3} - 6}\right] = x$

i.e. $\;$ $\dfrac{4^x + 1}{2^{x+3} - 6} = 2^x$

i.e. $\;$ $4^x + 1 = 2^x \left(2^{x+3} - 6\right)$

i.e. $\;$ $4^x + 1 = 2^{2x+3} - 6 \times 2^x$

i.e. $\;$ $\left(2^2\right)^x + 1 = \left(2^2\right)^x \times 2^3 - 6 \times 2^x$

i.e. $\;$ $8 \times \left(2^x\right)^2 - \left(2^x\right)^2 - 6 \times 2^x - 1 = 0$

i.e. $\;$ $7 \times \left(2^x\right)^2 - 6 \times 2^x - 1 = 0$

i.e. $\;$ $7 \times \left(2^x\right)^2 - 7 \times 2^x + 2^x - 1 = 0$

i.e. $\;$ $7 \times 2^x \left(2^x - 1\right) + 1 \left(2^x - 1\right) = 0$

i.e. $\;$ $\left(7 \times 2^x + 1\right) \left(2^x - 1\right) = 0$

i.e. $\;$ $2^x = \dfrac{-1}{7}$ $\;$ or $\;$ $2^x = 1$

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

Logarithim of a negative number is not defined.

$\therefore \;$ $x = \log_2 \left(\dfrac{-1}{7}\right)$ $\;$ is not a valid solution.

When $\;$ $x = \log_2 1$ $\implies$ $x = 0$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $1 - \dfrac{1}{2} \log \left(2x - 1\right) = \dfrac{1}{2} \log \left(x - 9\right)$

Given equation: $\;\;$ $1 - \dfrac{1}{2} \log \left(2x - 1\right) = \dfrac{1}{2} \log \left(x - 9\right)$

i.e. $\;$ $\dfrac{1}{2} \left[\log \left(x - 9\right) + \log \left(2x - 1\right)\right] = 1$

i.e. $\;$ $\log \left[\left(x - 9\right) \left(2x - 1\right)\right] = 2$ $\;\;\; \cdots \; (1)$

i.e. $\;$ $\left(x - 9\right) \left(2x - 1\right) = 10^2 = 100$

i.e. $\;$ $2 x^2 - 19 x + 9 = 100$

i.e. $\;$ $2 x^2 - 19x - 91 = 0$

i.e. $\;$ $\left(2x + 7\right) \left(x - 13\right) = 0$

i.e. $\;$ $x = \dfrac{-7}{2}$ $\;$ or $\;$ $x = 13$

When $\;$ $x = \dfrac{-7}{2}$, $\;$ the term $\;$ $\log \left(2x - 1\right)$ $\;$ in the given problem becomes

$\log \left[2 \times \left(\dfrac{-7}{2}\right) - 1\right] = \log \left(-8\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $x = \dfrac{-7}{2}$ $\;$ is not a valid solution.

When $\;$ $x = 13$, $\;$ the given problem [equation (1)] becomes

$\log \left[\left(13-9\right) \left(26-1\right) \right] = 2$

i.e. $\;$ $\log \left[4 \times 25\right] = 2$

i.e. $\log 100 = 2$ $\;$ which is true.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{13 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{\log \left(\sqrt{x + 1} + 1\right)}{\log \left(\sqrt[3]{x - 40}\right)} = 3$

Given equation: $\;\;$ $\dfrac{\log \left(\sqrt{x + 1} + 1\right)}{\log \left(\sqrt[3]{x - 40}\right)} = 3$

i.e. $\;$ $\log \left(\sqrt{x + 1} + 1\right) = 3 \log \left(\sqrt[3]{x - 40}\right)$

i.e. $\;$ $\log \left(\sqrt{x + 1} + 1\right) = \log \left(\sqrt[3]{x - 40}\right)^3$

i.e. $\;$ $\log \left(\sqrt{x + 1} + 1\right) = \log \left(x - 40\right)$ $\;\;\; \cdots \; (1)$

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

$\sqrt{x + 1} + 1 = x - 40$

i.e. $\;$ $\sqrt{x + 1} = x - 41$

i.e. $\;$ $x +1 = x^2 - 82x + 1681$

i.e. $\;$ $x^2 - 83x + 1680 = 0$

i.e. $\;$ $\left(x - 48\right) \left(x - 35\right) = 0$

i.e. $\;$ $x = 48$ $\;$ or $\;$ $x = 35$

When $\;$ $x = 35$, $\;$ the term $\;$ $\log \left(\sqrt[3]{x - 40}\right)$ $\;$ in the given problem becomes

$\log \left(\sqrt[3]{35 - 40}\right) = \log \left(\sqrt[3]{-5}\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ $x = 35$ $\;$ is not a valid solution.

When $\;$ $x = 48$, $\;$ the given problem becomes

$\dfrac{\log \left(\sqrt{48+1} + 1\right)}{\log \left(\sqrt[3]{48 - 40}\right)} = 3$

i.e. $\;$ $\dfrac{\log \left(\sqrt{49} + 1\right)}{\log \left(\sqrt[3]{8}\right)} = 3$

i.e. $\;$ $\dfrac{\log 8}{\log 2} = 3$

i.e. $\;$ $\dfrac{\log 2^3}{\log 2} = 3$

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

i.e. $\;$ $3 = 3$ $\;$ which is true.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{48 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) + 4 = 2x$

Given equation: $\;\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) + 4 = 2x$

i.e. $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) = 2x - 4$

i.e. $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) = 2\left(x - 2\right)$ $\;\;\; \cdots \; (1)$

Let $\;$ $x - 2 = p$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$\log_4 \left(2 \times 4^p - 1\right) = 2p$

i.e. $\;$ $2 \times 4^p - 1 = 4^{2p}$

i.e. $\;$ $\left(4^p\right)^2 - 2 \times 4^p + 1 = 0$

i.e. $\;$ $\left(4^p - 1\right)^2 = 0$

i.e. $\;$ $4^p - 1 = 0$

i.e. $\;$ $4^p = 1 = 4^0$

$\implies$ $p = 0$

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

$x - 2 = 0$ $\implies$ $x = 2$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$