Question

Solve the exponential equation. If the answer is not exact, give the answer to four decimal places. See Example 4. (Objective 1. Enter your answers as a comma-separated list.)

5^x^2=6^x

Answer 1

\[5^{x ^{2}} = 6^{x}\]

Answer 2

Taking the log to base 6 on both sides of the equation
\[\begin{align}
\log_6(5^{x^2})&=\log_6(6^x)\\
x^2\log_6(5)&=x\\
x^2\log_6(5)-x&=0\\
x\Big(x\log_6(5)-1\Big)&=0
\end{align}\]
A product of two terms equal zero.

The two solution can be found by setting one of terms in the product to zero.

\[x=0\] or \[x\log_65-1=0\]

Answer 3

the first solution was trivial,
but for the second, you will have to rearrange the equation a little ,and used the change of base formula

Answer 4

i dont understand why we plugged in base 6

Answer 5

\[x\log_6(5)-1=0\\
x\log_6(5)=1\\
x=\frac1{\log_6(5)}=\frac1{\log(5)/\log(6)}=\frac{\log(6)}{\log(5)}\]


They other way works too (maybe a little simpler)

Taking the log to base 5 on both sides of the equation
\[
\begin{align}
\log_5(5^{x^2})&=\log_5(6^x)\\
x^2&=x\log_5(6)\\
x^2-x\log_5(6)&=0\\
x(x-\log_5(6))&=0\\
\end{align}\]

\[x=0\]or\[x-\log_5(6)=0\\x=\log_5(6)\\x=\frac{\log(6)}{\log(5)}\]

Answer 6

The reason for taking the log to either base 5 or base 6, is to get rid of those exponentials

Answer 7

logarithm and exponential are kinda like inverses