Question

What is the inverse of y= 10^x +8?

Answer 1

`x = \log_(10)(y - 8)`, where
`y > 8`.

Explanation:

The question is asking for an expression for
`x` in terms of
`y`.

The first step is to move as many constants away from the side with
`x` as possible. In this question, that could be done by subtracting
`8` from both sides of the equation.

`y - 8 = 10^(x) + 8 - 8`.

`y - 8 = 10^(x)`.

Notice how
`x` is in the position of an exponent. Logarithms could help move
`x` out of that position.

For any base
`b > 0`:

`\log_(b) b^(x) = x\, \log_(b)\, b = x`.

In this question,
`10` would be the base. That is:
`b = 10`.

Take the logarithm (with
`10` as the base) of both sides of the equation
`y - 8 = 10^(x)`:

• Left-hand side:
  `\log_(10) (y - 8)`.
• Right-hand side:
  `\log_(10) \left(10^(x)\right) = x\, \log_(10)\left(10\right) = x`.

Equate both sides to find an expression for
`x`:

`x = \log_(10)(y - 8)`.