Question

How do you solve the exponential equation `100^(7x+1)=1000^(3x-2)`?

Answer 1

`x = -8/5`

Explanation:

In an exponential equation: `"base"^"exponent"`

Make both `100` and `1000` be a power of `10` so they both have the same `"base"`

`100 = 10^2` and `1000 = 10^3`

Use the exponent rule power of a power `(x^m)^n = x^(m*n)`

`(10^2)^(7x+1) = (10^3)^(3x-2)`

Simplify the exponents:

`10^(14x+2) = 10^(9x-6)`

When the bases are equivalent, the exponents are equivalent:
`14x+2 = 9x-6`

Solve for `x` by adding/subtracting like-terms:
`5x = -8`

So `x = -8/5`

Answer 2

I tried this:

Explanation:

Let us take the log in base `10` of both sides and use properties of logs:

`log_(10)100^(7x+1)=log _(10)1000^(3x-2)`

`(7x+1)log_(10)100=(3x-2)log_(10)1000`

`(7x+1)*2=(3x-2)*3`

`14x+2=9x-6`

`5x=-8`

`x=-8/5`