Question

What is the true solution to the equation below?

2 l n e Superscript l n 2 x Baseline minus l n e Superscript l n 10 x Baseline = l n 30
x = 30
x = 75
x = 150
x = 300

Answer 1

Interpret the expression as
2 ln(e^{ln(2x)}) − ln(e^{ln(10x)}) = ln 30.

Since ln(e^{A}) = A this becomes
2·ln(2x) − ln(10x) = ln 30.

Expand: 2(ln2 + ln x) − (ln10 + ln x) = ln30
⇒ ln x + 2ln2 − ln10 = ln30
⇒ ln x + ln(4) − ln(10) = ln30
⇒ ln[x·(4/10)] = ln30
⇒ x·(2/5) = 30
⇒ x = 30·5/2 = 75.

Check: 2 ln(150) − ln(750) = ln(150^2/750) = ln30. So the correct solution is x = 75.