Help me spot the error in my solution to the equation $25^x=6+5^x$

Question

I would use some help in figuring out where my math went wrong in solving the equation

$$25^x=6+5^x $$

My Approach:

$$ 5^{2x}-5^x = 6 $$ $$ 5^x(5^{2}-1) = 6 $$ $$ \ln(5^x(5^{2}-1)) = \ln(6) $$ $$ \ln(5^x) +\ln(5^{2}-1) = \ln(6) $$ $$ x\ln(5) +\ln(24) = \ln(6) $$ $$ x = \frac {\ln(6)-\ln(24)} {\ln(5)} $$ $$ x = \frac {\ln \frac{6}{24}} {\ln(5)} $$ $$ x = \frac {\ln \frac{3}{12}} {\ln(5)} =\log_5 \frac {3}{12}$$

The textbook solution of $25^x = 5^x + 6$ is but taking $25 = 5^2$ then noticing that we have a ‘quadratic in disguise’. Letting $u = 5^x$ we get $u^2 − u − 6 = 0$ which gives $u = −2$ or $u = 3$. Since $u = 5^x$ , we have $5^x = −2$ or $5^x = 3$. Since $5^x = −2$ has no real solution, we take natural logs and get $\ln (5^x ) = \ln(3)$ so that $x \ln(5) = \ln(3)$ or $x=\frac{\ln(3)}{\ln(5)}$.

Please note that I am not disputing the textbook solution. I just can't seem to figure out where I went wrong.

Answer 1

$5^{2x}-5^x\neq 5^x(5^2-1)$

We can see this quickly by letting $x=0$

Answer 2

$$5^{2x}-5^{x} = 6\implies \color{red}{5^x(5^2-1) = 6}$$ Your mistake was there. Recall the following two statements.

$$a^b\cdot a^c = a^{b+c} \neq \color\red{a^{bc}}$$ $$(a^b)^c = a^{bc}$$

Be sure not to confuse the two. So, instead of what you did, factor it like below.

$$5^x(5^x-1) = 6$$

Let $t = 5^x$ and solve for $t$.

Answer 3

Hint: write $$(5^x)^2=6+5^x$$ and Substitute $$5^x=t$$ then solve a quadratic. And $$5^{2x}-5^x\ne 5^x(5^2-1)=5^{x+2}-5^x$$

Answer 4

Note that $$5^{2x}-5^x = 6 \implies 5^x(5^x-1)=6$$

A better way to solve the problem is to let $y=5^x$ and solve for $y.$

Answer 5

The second line is not true. Try expanding it. You would notice it doesn't reproduce the first line. Rather, you can rewrite it in terms of base $5$ and rename $5^x$ as another variable which will lead to a quadratic equation.