How do you solve ${7^x} = 80$?

Hint:In this question, we need to solve ${7^x} = 80$. Here, we will determine, between which value where $80$ exists. Then, we will convert the given from exponential form to log form. Then, we will use the base rule. Then, using the log table we will find the required values and substitute to determine the required value of $x$.

Complete step-by-step solution:
In this question we need to solve ${7^x} = 80$.

Now, let’s substitute the value of $x$ from $0$, to determine, between which value $80$ exists.

Thus, substituting $x = 0$, we have,
${7^0} = 1$

By substituting $x = 1$, we have,
${7^1} = 7$

By substituting $x = 2$, we have,
${7^2} = 49$

Now by substituting $x = 3$, we have,
${7^3} = 343$

Therefore, it is clear that $80$ lies between $49$ and $343$ i.e., between ${7^2}$ and ${7^3}$.

In order to determine the value of $x$, we are now converting the exponent form into log form.

Therefore, we can rewrite the equation as $x = {\log _7}80$

We can calculate the value by using the calculator or we can use the logarithm change of base rule.
The logarithm change of base rule states that, \[{\text{lo}}{{\text{g}}_b}\left( x \right){\text{ =
}}\dfrac{{{\text{lo}}{{\text{g}}_c}\left( x \right)}}{{{\text{lo}}{{\text{g}}_c}\left( b \right)}}{\text{ }}\]

Hence, $x = \dfrac{{\log 80}}{{\log 7}}$

Now, we will take the values of $\log 80$ and $\log 7$ from the logarithm table. Therefore,

log 80 = 1.903
log 7 = 0.845

Therefore, by substituting the values, we get,
$x = \dfrac{{1.903}}{{0.845}}$

$x = 2.25$

Hence, the value of $x$ in ${7^x}$ = 80 is $2.25$ (approx.)

Note: In this question it is important to note that to change from exponential form to logarithmic form, identify the base of the exponential equation and move the base to the other side of the equal sign and add the word “log”. Do not move anything but the base, the other numbers or variables will not change. Also, logarithms are the inverse of exponential function.