Solve 4^x = 80 for x.

Step-by-step solution

1. 1

   Check whether $80$ is a power of $4$. The powers are $4^1=4$, $4^2=16$, $4^3=64$, $4^4=256$. Since $64<80<256$, the exponent is not an integer and lies strictly between $3$ and $4$ — worth noting now as a bracket for the final answer.

2. 2

   Take logarithms of both sides. Any base works; the logarithm turns the exponent into a factor:

   $\ln\left(4^x\right)=\ln 80\quad\Longrightarrow\quad x\ln 4=\ln 80$

3. 3

   Solve for $x$ and name the exact form.

   $x=\frac{\ln 80}{\ln 4}=\log_4 80$

   This is the change-of-base formula read backwards: $\log_b a=\tfrac{\ln a}{\ln b}$.

4. 4

   Evaluate numerically. With $\ln 80\approx 4.3820266$ and $\ln 4\approx 1.3862944$:

   $x\approx\frac{4.3820266}{1.3862944}\approx 3.16096$

5. 5

   Check against the bracket and by substitution. The value $3.161$ does lie between $3$ and $4$ $\checkmark$, and $4^{3.16096}\approx 80.000$ $\checkmark$. An exact alternative form: since $80=16\times 5$ and $16=4^2$, we get $x=2+\log_4 5$, and $\log_4 5\approx 1.16096$ $\checkmark$.