Answer:
[tex]x = \log_{10}(y - 8)[/tex], where [tex]y > 8[/tex].
Step-by-step explanation:
The question is asking for an expression for [tex]x[/tex] in terms of [tex]y[/tex].
The first step is to move as many constants away from the side with [tex]x[/tex] as possible. In this question, that could be done by subtracting [tex]8[/tex] from both sides of the equation.
[tex]y - 8 = 10^{x} + 8 - 8[/tex].
[tex]y - 8 = 10^{x}[/tex].
Notice how [tex]x[/tex] is in the position of an exponent. Logarithms could help move [tex]x[/tex] out of that position.
For any base [tex]b > 0[/tex]:
[tex]\log_{b} b^{x} = x\, \log_{b}\, b = x[/tex].
In this question, [tex]10[/tex] would be the base. That is: [tex]b = 10[/tex].
Take the logarithm (with [tex]10[/tex] as the base) of both sides of the equation [tex]y - 8 = 10^{x}[/tex]:
Equate both sides to find an expression for [tex]x[/tex]:
[tex]x = \log_{10}(y - 8)[/tex].
