Question

How do you solve `81^ { 2x + 3} = 243^ { x + 2}`?

Answer 1

The Real solution is:

`x = -2/3`

Complex solutions are all of the form:

`x = -2/3+(2kpii)/(3ln3)" "` where `k` is any integer

Explanation:

Note that `81 = 3^4` and `243 = 3^5`, so...

`3^(4(2x+3)) = (3^4)^(2x+3) = 81^(2x+3) = 243^(x+2) = (3^5)^(x+2) = 3^(5(x+2))`

The function `f(t) = 3^t` is one to one as a Real valued function of Real numbers.

So given:

`3^(4(2x+3)) = 3^(5(x+2))`

any Real solution satisfies:

`4(2x+3) = 5(x+2)" "color(blue)(tt"(i)")`

Multiplied out:

`8x+12 = 5x+10`

Subtract `5x+12` from both sides to get:

`3x=-2`

So (dividing both sides by `3`), we find:

`x = -2/3`

This is the unique Real solution of the original equation.

`color(white)()`
Complex solutions

Note that:

`e^(2kpii) = 1`

for any integer value of `k`

Also:

`3^t = (e^(ln3))^t = e^(tln 3)`

Hence:

`3^((2kpii)/ln 3) = e^(2kpii) = 1`

So we can modify `color(blue)(tt"(i)")` from above and assert:

`4(2x+3) = 5(x+2) + (2kpii)/(ln3)" "color(blue)(tt"(ii)")`

Hence:

`3x = -2+ (2kpii)/(ln3)`

So:

`x = -2/3+(2kpii)/(3ln3)`