Why do I receive the wrong answer when I try to solve this exponential equation?

Question

So I have the equation: $25^{x}=5^{x}+6$

My reasoning is if you make everything to the base 5:

$\left( 5^{2}\right) ^{x}=5^{x}+5^{\log _{5}6}$

Given the bases are the same we can do:

$2x=x+\log _{5}6$

$x=\log _{5}6$

---

This answer is wrong however, why is this? Once I've the same bases why can't I do this? Furthermore why couldn't I just take logs of both sides of the original equation?($25^{x}=5^{x}+6$). What law of logarithms stops me from doing this, why?

Thank you.

Answer 1

You can't just 'distribute' the $\log$ function over addition. To solve this, notice that $$ 25^x=5^x+6 $$ is really just a quadratic equation in disguise $$ 5^{2x}-5^x-6=0 $$ So let $u=5^x$, then we have $$ u^2-u-6=(u-3)(u+2)=0 $$ Which I'm sure you can easily solve. Once you get the solutions for $u$, it is simple to get the solutions for $x$ using $5^x=u$.

Answer 2

If $y=5^x$, you have $y^2-y-6=(y-3)(y+2)=0$ so you should be able to solve for $y$ and then $x$.

Answer 3

As $25=5^2, 25^x=(5^2)^x=(5^x)^2$

$$25^x=5^x+6\implies (5^x)^2-(5^x)-6=0$$

$$\implies 5^x=\frac{1\pm\sqrt{25}}2=3,-2$$

If $x$ is real, $5^x>0\implies 5^x=3\implies x=\log_53$