Question

Match each logarithmic equation to its equivalent exponential form. The logarithmic equations are in(5x+1) =in(2x+10), log_4(3x-2) =log_4(x+6), log(x-10) =log(3x-12), log_7(x+5) =log_7(4x-1). The equivalent exponential forms are x=3, no solution, x=4, x=2.

Answer 1

Matchings with brief work:

- ln(5x+1) = ln(2x+10) → 5x+1 = 2x+10 ⇒ 3x = 9 ⇒ x = 3.
- log_4(3x-2) = log_4(x+6) → 3x-2 = x+6 ⇒ 2x = 8 ⇒ x = 4.
- log(x-10) = log(3x-12) → x-10 = 3x-12 ⇒ 2 = 2x ⇒ x = 1, but x must be >10, so no solution.
- log_7(x+5) = log_7(4x-1) → x+5 = 4x-1 ⇒ 3x = 6 ⇒ x = 2.

So: ln(5x+1)=ln(2x+10) → x=3; log_4(3x-2)=log_4(x+6) → x=4; log(x-10)=log(3x-12) → no solution; log_7(x+5)=log_7(4x-1) → x=2.