Solving $(\log_2 x) - 4 = 5\log_x 2$

Question

How do I solve the following equation?

$$(\log_2 x ) - 4 = 5\log_x 2$$

I tried doing this:

log (x * X/32) = 4 2 and i got x^2 = 512

Also, how do I solve equations of the form log⋅log⋅=… ? logarithms

Answer 1

HINT:

For use $$\log_2x-4=5\log_x2$$

use $\log_ab=\dfrac{\log b}{\log a}=\dfrac1{\log_ba}$

Answer 2

Note:

$$ \log_b a = \frac{1}{\log_a b}$$

Hence your equation becomes:

$$ (\log_2 x) - 4 = \frac{5}{\log_2 x}$$

Let $\log_2 x$ be $n$

$$ n - 4 =\frac{5}{n}$$

$$ n^2 -4n-5= 0$$

Solve the quadratic above and then substitute back into the following equation to find $x$:

$\log_2 x = n$

You should obtain two values for $x$

Note that:

$$ \log_b a = k \ \implies \ a= b^k$$

Answer 3

Rewrite $$5\log_x 2 = 5\frac{\log_22}{\log_2x},$$ then rearrange the terms to get $$(\log_2x)^2 - 4\log_2x - 5 = 0$$ and make the substitution $$y = \log_2x$$ and solve the quadratic $$y^2 - 4y - 5 = 0.$$

Answer 4

Notice, logarithms is defined for positive real numbers, now we have $$\log_{2}x-4=5\log_x2$$ we have following restrictions $x>0$ & $(\log_2x)\neq 0$ or $x\neq 1$

$$\log_{2}x-4=\frac{5}{\log_2x}$$ $$(\log_2x)^2-4\log_2x=5$$ $$(\log_2x)^2-4\log_2x-5=0$$ $$(\log_2x)^2-4\log_2x-5=0$$ $$\underbrace{(\log_2x)^2-5\log_2x}+\underbrace{\log_2x-5}=0$$ $$\log_2x(\log_2x-5)+(\log_2x+5)=0$$ $$(\log_2x-5)(\log_2x+1)=0$$ Now, solving for $x$ $$\log_2x-5=0\implies \log_2x=5\implies x=2^5=\color{red}{32}$$ or
$$\log_2x+1=0\implies \log_2x=-1\implies x=2^{-1}=\color{red}{\frac{1}{2}}$$