Question

2. Solve 3^4x-5 = (1/27)^2x+10 for x. Show your work

I know it equals around the negatives but that's it

Answer 1

It looks like the equation is supposed to read

3⁴ˣ ⁻ ⁵ = (1/27)²ˣ ⁺ ¹⁰

Recall that 3³ = 27, so 1/27 = 3⁻³ :

3⁴ˣ ⁻ ⁵ = (3⁻³)²ˣ ⁺ ¹⁰

Distribute the exponent on the right side:

3⁴ˣ ⁻ ⁵ = 3 ⁻⁶ˣ ⁻ ³⁰

The exponents must be equal for this equation to hold, so

4x - 5 = -6x - 30

Solve for x :

10x = -25

x = -25/10 = -5/2