Solve $\log_2x= \log_4(x+6)$

Question

Solve $\log_2x= \log_4(x+6)$ for $x$ using the change of base formula.

I already tried changing the base on both sides but that didn't work I know it must be in the form of a quadratic for a substitution to be made.

Answer 1

Hint:

Assume: $$\log_2x=y$$ then $$x=2^y=(4^{1/2})^y=4^{y/2}$$

which means $$\log_4 x=y/2$$

Answer 2

Hint:

From the change of basis formula you have $ \log_2 x=2\log_4 x=\log_4 x^2$

and $$ \log_4 x^2=\log_4 (x+6) $$ in $\mathbb{R}$, iff

$$ x^2=(x+6) \quad \land \quad x+6>0 $$

Answer 3

or you can write $$\frac{\ln(x)}{\ln(2)}=\frac{\ln(x+6)}{2\ln(2)}$$ i think this will help you