Question

What are the solutions of `100^(7x+1) = 1000^(3x-2)` ?

Answer 1

Real valued solution:

`x=-8/5`

Complex valued solutions:

`x = -8/5+(2kpii)/(5ln(10))" "` for any integer `k`

Explanation:

Note that if `a > 0` then:

`(a^b)^c = a^(bc)`

So we find:

`10^(14x+2) = 10^(2(7x+1)) = (10^2)^(7x+1) = 100^(7x+1)`

`10^(9x-6) = 10^(3(3x-2)) = (10^3)^(3x-2) = 1000^(3x-2)`

So if:

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

then:

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

Then either say "take common logarithms of both sides" or simply note that as a real-valued function `10^x` is one to one to deduce:

`14x+2 = 9x-6`

Subtract `9x+2` from both sides to get:

`5x=-8`

Divide both sides by `5` to get:

`x = -8/5`

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Footnote

There are also Complex valued solutions, which we can deduce from Euler's identity:

`e^(ipi)+1 = 0`

Hence:

`e^(2kpii) = 1" "` for any integer `k`

Then:

`10 = e^(ln 10)`

So:

`10^x = e^(x ln 10)`

So we find:

`10^((2kpii)/ln(10)) = 1" "` for any integer `k`

Hence we find:

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

if and only if:

`14x+2 = 9x-6 + (2kpii)/ln(10)" "` for some integer `k`

Hence:

`5x = -8+(2kpii)/ln(10)`

So:

`x = -8/5+(2kpii)/(5ln(10))`