Algebra - Logarithmic Equations

Solve the equation: $\;$ $\left[\sqrt{x}\right]^{\left(\log_5 x\right) - 1} = 5$

Given equation: $\;\;$ $\left[\sqrt{x}\right]^{\left(\log_5 x\right) - 1} = 5$

i.e. $\;$ $\left(\sqrt{x}\right)^{\log_5 x} \times \left(\sqrt{x}\right)^{-1} = 5$

i.e. $\;$ $\dfrac{\left(\sqrt{x}\right)^{\log_5 x}}{\sqrt{x}} = 5$ $\;\;\; \cdots \; (1)$

Taking logarithims to base $5$ on both sides of equation $(1)$ gives

$\log_5 \left[\dfrac{\left(\sqrt{x}\right)^{\log_5 x}}{\sqrt{x}}\right] = \log_5 5$

i.e. $\;$ $\log_5 \left(\sqrt{x}\right)^{\log_5 x} - \log_5 \sqrt{x} = 1$

i.e. $\;$ $\log_5 x \times \log_5 \sqrt{x} - \log_5 \sqrt{x} = 1$

i.e. $\;$ $\log_5 x \times \log_5 x^{\frac{1}{2}} - \log_5 x^{\frac{1}{2}} = 1$

i.e. $\;$ $\dfrac{1}{2} \log_5 x \times \log_5 x - \dfrac{1}{2} \log_5 x = 1$

i.e. $\;$ $\left(\log_5 x\right)^2 - \log_5 x - 2 = 0$

i.e. $\;$ $\left(\log_5 x\right)^2 - 2 \log_5 x + \log_5 x - 2 = 0$

i.e. $\;$ $\log_5 x \left(\log_5 x - 2\right) + 1 \left(\log_5 x - 2\right) = 0$

i.e. $\;$ $\left(\log_5 x + 1\right) \left(\log_5 x - 2\right) = 0$

i.e. $\;$ $\log_5 x = -1$ $\;$ or $\;$ $\log_5 x = 2$

$\implies$ $x = 5^{-1} = \dfrac{1}{5}$, $\;$ or $\;$ $x = 5^2 = 25$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{5}, \; 25 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\left(\log x\right) + 1} = 10^6$

Given equation: $\;\;$ $x^{\left(\log x\right) + 1} = 10^6$

i.e. $\;$ $x^{\log x} \times x^1 = 10^6$ $\;\;\; \cdots \; (1)$

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

$\log \left(x^{\log x} \times x\right) = \log {10}^6$

i.e. $\;$ $\log \left(x^{\log x}\right) + \log x = 6 \log 10$

i.e. $\;$ $\log x \times \log x + \log x = 6$

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

i.e. $\;$ $\left(\log x\right)^2 + 3 \log x - 2 \log x - 6 = 0$

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

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

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

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

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\log_3 x} = 9$

Given equation: $\;\;$ $x^{\log_3 x} = 9$ $\;\;\; \cdots \; (1)$

Taking logarithims to the base $3$ on both sides of equation $(1)$ gives

$\log_3 \left(x^{\log_3 x}\right) = \log_3 9$

i.e. $\;$ $\log_3 x \times \log_3 x = \log_3 3^2$

i.e. $\;$ $\left(\log_3 x\right)^2 = 2 \log_3 3$

i.e. $\;$ $\left(\log_3 x\right)^2 = 2$

i.e. $\;$ $\log_3 x = \pm \sqrt{2}$

i.e. $\;$ $x = 3^{\pm \sqrt{2}}$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\frac{\log x + 5}{3}} = 10^{5 + \log x}$

Given equation: $\;\;$ $x^{\frac{\log x + 5}{3}} = 10^{5 + \log x}$ $\;\;\; \cdots \; (1)$

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

$\log \left(x^{\frac{\log x + 5}{3}}\right) = \log \left(10^{5 + \log x}\right)$

i.e. $\;$ $\left(\dfrac{\log x + 5}{3}\right) \log x = \left(5 + \log x\right) \log 10$

i.e. $\;$ $\dfrac{1}{3} \left(\log x\right)^2 + \dfrac{5}{3} \log x = 5 + \log x$

i.e. $\;$ $\dfrac{1}{3} \left(\log x\right)^2 + \dfrac{2}{3} \log x - 5 = 0$

i.e. $\;$ $\left(\log x\right)^2 + 2 \log x - 15 = 0$

i.e. $\;$ $\left(\log x\right)^2 + 5 \log x - 3 \log x - 15 = 0$

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

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

i.e. $\;$ $\log x = 3$ $\;\;$ or $\;\;$ $\log x = -5$

$\implies$ $x = 10^3$ $\;\;$ or $\;\;$ $x = 10^{-5}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-5}, \; 10^3 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{2 \log x} = 10 x^2$

Given equation: $\;\;$ $x^{2 \log x} = 10 x^2$ $\;\;\; \cdots \; (1)$

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

$\log \left(x^{2 \log x}\right) = \log \left(10 x^2\right)$

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

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

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

i.e. $\;$ $\log x = \dfrac{2 \pm \sqrt{4 + 8}}{4} = \dfrac{2 \pm \sqrt{12}}{4}$

i.e. $\;$ $\log x = \dfrac{2 \pm 2 \sqrt{3}}{4}$

i.e. $\;$ $\log x = \dfrac{1 \pm \sqrt{3}}{2}$

$\implies$ $x = 10^{\frac{1 \pm \sqrt{3}}{2}}$

i.e. $\;$ $x = \left(10^{1 \pm \sqrt{3}}\right)^{\frac{1}{2}}$

i.e. $\;$ $x = \sqrt{10^{1 \pm \sqrt{3}}}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\sqrt{10^{1 + \sqrt{3}}}, \; \sqrt{10^{1 - \sqrt{3}}} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $9^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$

Given equation: $\;\;$ $9^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$

i.e. $\;$ $\left(3^2\right)^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$

i.e. $\;$ $\left(3\right)^{2 \log_3 \left(1 - 2x\right)} = 5x^2 - 5$

i.e. $\;$ $3^{\log_3 \left(1 - 2x\right)^2} = 5x^2 - 5$

i.e. $\;$ $\left(1 - 2x\right)^2 = 5x^2 - 5$

i.e. $\;$ $1 - 4x + 4x^2 = 5x^2 - 5$

i.e. $\;$ $x^2 + 4x - 6 = 0$

i.e. $\;$ $x = \dfrac{-4 \pm \sqrt{16 + 24}}{2}$

i.e. $\;$ $x = \dfrac{-4 \pm \sqrt{40}}{2}$

i.e. $\;$ $\dfrac{-4 \pm 2 \sqrt{10}}{2}$

i.e. $\;$ $x = -2 \pm \sqrt{10}$

Now, $\;$ $\sqrt{10} \approx 3.16, \;\;\; -2 + \sqrt{10} = 1.16$

and $\;$ $\log_3 \left(1 - 2x\right) = \log_3 \left(1 - 2.32\right) = \log_3 \left(-1.32\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $x = -2 + \sqrt{10}$ $\;$ is not a valid solution.

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{1 + \log x} = 10x$

Given equation: $\;\;$ $x^{1 + \log x} = 10x$

i.e. $\;$ $1 + \log x = \log_x {10x}$

i.e. $\;$ $1 + \log x = \log_x {10} + \log_x x$ $\;\;\;$ $\left[\because \; \log_a \left(CD\right) = \log_a C + \log_a D\right]$

i.e. $\;$ $1 + \log x = \dfrac{\log_{10} 10}{\log_{10} x} + 1$ $\;\;\;$ $\left[\because \; \log_b a = \dfrac{\log_m a}{\log_m b}; \;\; \log_a a = 1\right]$

i.e. $\;$ $\log_{10} x = \dfrac{1}{\log_{10} x}$

i.e. $\;$ $\left(\log_{10} x\right)^2 = 1$

i.e. $\;$ $\log_{10} x = \pm 1$

Now, $\;$ $\log_{10} x = 1$ $\implies$ $x = 10^1 = 10$

and $\;$ $\log_{10} x = -1$ $\implies$ $x = 10^{-1} = \dfrac{1}{10} = 0.1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0.1, \; 10 \right\}$