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

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.

• HINT: Let $y=5^x$ to give $y^2=y+6$ and solve this quadratic – Mufasa Dec 1 '13 at 15:00
• $\log(a+b) \neq \log(a) + \log(b)$, so taking logs on both sides doesn't help. – Macavity Dec 1 '13 at 15:02

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 $5^x=u$.

• I'm aware of this method of solving but I wonder why you can't just equate the powers as both sides have exponentials with a base 5? – seeker Dec 1 '13 at 16:06
• @Assad Because it's not something like $25^x=5^{3x+1}$ which you can rewrite as $5^{2x}=5^{3x+1}$, where you would then equate the powers. In your problem, you have addition in there. What you are really doing is saying, 'Hey, if both these sides are equal then if I plug them into $\log$ I should get the same thing.' Then $\log_5 25^x=\log_5(5^x+6)$. The problem is that is not the same thing as $\log_5 25^x=\log_5 5^x + \log_5 6$. That is because log's don't have the property that $\log(a+b)=\log a +\log b$. They do have the property $\log(ab)=\log a+\log b$. – mathematics2x2life Dec 1 '13 at 16:12
• @Assad So it depends on what you start with whether you would turn it into a quadratic equation, relate the powers, or solving by taking the inverse of both sides. But don't worry, after some practice it'll be really obvious to you which you'll need to do! – mathematics2x2life Dec 1 '13 at 16:13
• Oh okay! Thanks a lot! – seeker Dec 1 '13 at 16:20

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 $x$.

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