Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_4 \left(\log_2 x\right) + \log_2 \left(\log_4 x\right) = 2$

Given equation: $\;\;$ $\log_4 \left(\log_2 x\right) + \log_2 \left(\log_4 x\right) = 2$ $\;\;\; \cdots \; (1)$

Now, $\;$ $\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \dfrac{\log_2 x}{\log_2 2^2} = \dfrac{\log_2 x}{2 \log_2 2}$

i.e. $\;$ $\log_4 x = \dfrac{\log_2 x}{2}$ $\;\;\; \cdots \; (2)$

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

$\log_4 \left(\log_2 x\right) + \dfrac{1}{2} \log_2 \left(\log_2 x\right) = 2$ $\;\;\; \cdots \; (3)$

Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (4)$

Then equation $(3)$ becomes

$\log_4 p + \dfrac{1}{2} \log_2 p = 2$

i.e. $\;$ $2 \log_4 p + \log_2 p = 4$

i.e. $\;$ $2 \times \dfrac{\log_2 p}{\log_2 4} + \log_2 p = 4$

i.e. $\;$ $2 \times \dfrac{\log_2 p}{\log_2 2^2} + \log_2 p = 4$

i.e. $\;$ $2 \times \dfrac{\log_2 p}{2 \log_2 2} + \log_2 p = 4$

i.e. $\;$ $\log_2 p + \log_2 p = 4$

i.e. $\;$ $2 \log_2 p = 4$

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

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

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

$\log_2 x = 4$ $\implies$ $x = 2^4 = 16$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{16 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{3 \log x - \frac{1}{\log x}} = \sqrt[3]{10}$

Given equation: $\;\;$ $x^{3 \log x - \frac{1}{\log x}} = \sqrt[3]{10}$

i.e. $\;$ $\left(x^{3 \log x - \frac{1}{\log x}}\right)^3 = \left(10^{\frac{1}{3}}\right)^3$

i.e. $\;$ $x^{9 \log x - \frac{3}{\log x}} = 10$ $\;\;\; \cdots \; (1)$

Taking logarithim to base $10$ on both sides of equation $(1)$ gives

i.e. $\;$ $\log \left(x^{9 \log x - \frac{3}{\log x}}\right) = \log 10$

i.e. $\;$ $\left(9 \log x - \frac{3}{\log x}\right) \log x = 1$

i.e. $\;$ $9 \left(\log x\right)^2 - 3 = 1$

i.e. $\;$ $9 \left(\log x\right)^2 = 4$

i.e. $\;$ $3 \log x = \pm 2$

i.e. $\;$ $\log x = \dfrac{2}{3}$, $\;$ or $\;$ $\log x = \dfrac{-2}{3}$

$\implies$ $x = 10^{\frac{2}{3}}$ $\;$ or $\;$ $x = 10^{\frac{-2}{3}}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{\frac{-2}{3}}, \; 10^{\frac{2}{3}} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(x^3 + 27\right) - 0.5 \log \left(x^2 + 6x + 9\right) = 3 \log \sqrt[3]{7}$

Given equation: $\;\;$ $\log \left(x^3 + 27\right) - 0.5 \log \left(x^2 + 6x + 9\right) = 3 \log \sqrt[3]{7}$

i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 -3x + 9\right)\right] - 0.5 \log \left(x + 3\right)^2 = \log \left(\sqrt[3]{7}\right)^3$

i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 - 3x + 9\right)\right] - 0.5 \times 2 \log \left(x + 3\right) = \log 7$

i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 - 3x + 9\right)\right] - \log \left(x + 3\right) = \log 7$

i.e. $\;$ $\log \left[\dfrac{\left(x + 3\right) \left(x^2 - 3x + 9\right)}{x + 3}\right] = \log 7$

i.e. $\;$ $\log \left(x^2 - 3x + 9\right) = \log 7$ $\;\;\; \cdots \; (1)$

Taking antilog to base $10$ on both sides of equation $(1)$ gives

$x^2 - 3x + 9 = 7$

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

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

i.e. $\;$ $x = 2$ $\;$ or $\;$ $x = 1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1, \; 2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(2x^2\right) \times \log_2 \left(16 x\right) = \dfrac{9}{2} \log^2_2 x$

Given equation: $\;\;$ $\log_2 \left(2x^2\right) \times \log_2 \left(16 x\right) = \dfrac{9}{2} \log^2_2 x$

i.e. $\;$ $\left(\log_2 2 + \log_2 x^2\right) \times \left(\log_2 16 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$

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

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

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

Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$\left(1 + 2p\right) \left(4 + p\right) = \dfrac{9}{2} p^2$

i.e. $\;$ $4 + 9p + 2p^2 = \dfrac{9}{2} p^2$

i.e. $\;$ $8 + 18 p + 4p^2 = 9 p^2$

i.e. $\;$ $5p^2 - 18p - 8 = 0$

i.e. $\;$ $\left(5p + 2\right) \left(p - 4\right) = 0$

i.e. $\;$ $5p + 2 = 0$ $\;$ or $\;$ $p - 4 = 0$

i.e. $\;$ $p = \dfrac{-2}{5}$, $\;$ or $\;$ $p = 4$

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

when $\;$ $p = \dfrac{-2}{5}$, $\;$ we have from equation $(2)$

$\log_2 x = \dfrac{-2}{5}$ $\implies$ $x = 2^{\frac{-2}{5}}$

when $\;$ $p = 4$, $\;$ we have from equation $(2)$

$\log_2 x = 4$ $\implies$ $x = 2^4 = 16$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2^{\frac{-2}{5}}, \; 16 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(4^{x+1} + 4\right) \times \log_2 \left(4^x + 1\right) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\dfrac{1}{8}}$

Given equation: $\;\;$ $\log_2 \left(4^{x+1} + 4\right) \times \log_2 \left(4^x + 1\right) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\dfrac{1}{8}}$

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

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

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

i.e. $\;$ $\log_2 2^2 \times \log_2 \left(4^x + 1\right) + \left[\log_2 \left(4^x + 1\right)\right]^2 = 3$

i.e. $\;$ $\left[\log_2 \left(4^x + 1\right)\right]^2 + 2 \log_2 \left(4^x + 1\right) - 3 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\log_2 \left(4^x + 1\right) = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$p^2 + 2p - 3 = 0$

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

i.e. $\;$ $p = -3$ $\;$ or $\;$ $p = 1$

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

when $\;$ $p = -3$,

$\log_2 \left(4^x + 1\right) = -3$

i.e. $\;$ $4^x + 1 = 2^{-3} = \dfrac{1}{8}$

i.e. $\;$ $4^x = \dfrac{-7}{8}$ $\implies$ $x = \log_4 \left(\dfrac{-7}{8}\right)$

But, logarithim of a negative number is not defined.

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

when $\;$ $p = 1$,

$\log_2 \left(4^x + 1\right) = 1$

i.e. $\;$ $4^x + 1 = 2^1 = 2$

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

i.e. $\;$ $x = \log_4 1$ $\implies$ $x = 0$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log 2 + \log \left(4^{x-2} + 9\right) = 1 + \log \left(2^{x-2} + 1\right)$

Given equation: $\;\;$ $\log 2 + \log \left(4^{x-2} + 9\right) = 1 + \log \left(2^{x-2} + 1\right)$

i.e. $\;$ $\log \left(4^{x-2} + 9\right) - \log \left(2^{x-2} + 1\right) = 1 - \log 2$

i.e. $\;$ $\log \left[\dfrac{4^{x-2} + 9}{2^{x-2} + 1}\right] = \log 10 - \log2$

i.e. $\;$ $\log \left[\dfrac{\left(2^2\right)^{x-2} + 9}{2^{x-2} + 1}\right] = \log \left(\dfrac{10}{2}\right)$

i.e. $\;$ $\log \left[\dfrac{2^{2x-4} + 9}{2^{x-2} + 1}\right] = \log 5$ $\;\;\; \cdots \; (1)$

Taking antilog to base $10$ on both sides of equation $(1)$ gives

$\dfrac{2^{2x-4} + 9}{2^{x-2} + 1} = 5$

i.e. $\;$ $2^{2x-4} + 9 = 5 \times 2^{x-2} + 5$

i.e. $\;$ $\dfrac{2^{2x}}{2^4} - 5 \times \dfrac{2^x}{2^2} + 4 = 0$

i.e. $\;$ $2^{2x} - 5 \times 2^x \times 2^2 + 4 \times 2^4 = 0$

i.e. $\;$ $\left(2^x\right)^2 - 20 \times 20^x + 64 = 0$

i.e. $\;$ $\left(2^x\right)^2 - 16 \times 2^x - 4 \times 2^x + 64 = 0$

i.e. $\;$ $2^x \left(2^x - 16\right) - 4 \left(2^x - 16\right) = 0$

i.e. $\;$ $\left(2^x - 4\right) \left(2^x - 16\right) = 0$

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

i.e. $\;$ $2^x = 4 = 2^2$ $\;$ or $\;$ $2^x = 16 = 2^4$

$\implies$ $x = 2$ $\;$ or $\;$ $x = 4$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 4 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left[\sqrt{130 - 7^{\log_x \left(6 - x\right)}}\right] = 2$

Given equation: $\;\;$ $\log_3 \left[\sqrt{130 - 7^{\log_x \left(6 - x\right)}}\right] = 2$

i.e. $\;$ $\dfrac{1}{2} \log_3 \left[130 - 7^{\log_x \left(6 - x\right)}\right] = 2$

i.e. $\;$ $\log_3 \left[130 - 7^{\log_x \left(6 - x\right)}\right] = 4$

i.e. $\;$ $130 - 7^{\log_x \left(6 - x\right)} = 3^4 = 81$

i.e. $\;$ $7^{\log_x \left(6 - x\right)} = 49 = 7^2$

i.e. $\;$ $\log_x \left(6 - x\right) = 2$

i.e. $\;$ $6 - x = x^2$

i.e. $\;$ $x^2 + x - 6 = 0$

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

i.e. $\;$ $x = -3$ $\;$ or $\;$ $x = 2$

But $\;$ $x = -3$ $\;$ makes the base of the term $\;$ $\log_x \left(6 - x\right)$, $\;$ in the given question, negative.

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

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