Question

How do you solve `3^(4x) = 3^(5-x)`?

Answer 1

`x=1`

Explanation:

`color(blue)("Quickest way using shortcuts")`
`color(brown)("If you have "x^a=x^b" then "a=b)`

`=>4x=5-x`

`4x+x=5`

`5x=5 => x=1`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Another method that demonstrates other properties of indices.")`

Note that `3^(5-x)" is the same as "3^5/(3^x`

Write as `3^(4x)=3^5/3^x`

Multiply both sides by `3^x`

`3^(4x)xx3^x=3^5 xx3^x/3^x`

But `3^x/3^x=1" and "3^(4x)xx3^x=3^(5x)` giving

`3^(5x)=3^5`

Comparing just the indices

`5x=5`

`=>x=1`

Answer 2

Notice from the outset that the two exponents must be equal, since their bases are both `3`.

`color(red)3^color(blue)(4x)=color(red)3^color(blue)(5-x)" "=>" "color(blue)(4x)=color(blue)(5-x)" "=>" "5x=5`

Thus, `x=1`.

Answer 3

`x=1`

Explanation:

As a Real-valued function of Real numbers, `f(x) = 3^x` is strictly monotonically increasing and is therefore one-to-one.

So `f(4x) = 3^(4x) = 3^(5-x) = f(5-x)` implies `4x = 5-x`

Add `x` to both sides to get:

`5x=5`

Divide both sides by `5` to get:

`x=1`