# Is this logarithm-based way to solve an exponential equation valid?

A friend of mine solved the following exponential equation in the following way:

$$4^{x+1}-8^{2x}=0$$

Since $\log_{4}8=\frac{3}{2}$, $4^{x+1}-(4^{2x})^{\frac{3}{2}})=0 \Leftrightarrow 4^{x+1}-4^{3x}=0$. Now that both $x$-terms are base 4, this equation is trivial to solve. However, I don't understand how that process can work. Is there any base to this process or was it just a lucky coincidence?

Note: I know how to solve that equation, I'm only asking if my friend's approach is valid or not

• Yes, your friend's approach is valid – Omnomnomnom Jan 30 '17 at 19:45
• @Omnomnomnom Interesting, can you point me to some resources/formulas for this? – imas145 Jan 30 '17 at 19:51

Well you have that $8=2^{3}=(\sqrt{4})^3=4^{3/2}$.Now as to why it works we have that $$a^{\log_a b}=b$$ Now plugging in $b=8$ we can write $$8=4^{\log_4 8}$$ And we have that $\log_4 8=\frac{3}{2}$ So we can write $8^{2x}$ as $(4^{3/2})^{2x}$. Your friends approach is correct.
Another, less convoluted way is to convert everything to base 2: $$0 = (2^2)^{x+1} - (2^3)^{2x} = 2^{2x+2} - 2^{6x}...$$