LOG log log

use the equality:
ln(xy) = ln(x) + ln(y)

then:
ln(3x) + ln(5x-2) = 3
ln(3x*(5x-2)) = 3
3x*(5x-2) = e^3

by solving the above quadratic equation, we have:
x = 1.37 or x = -0.97

but you need to check that x should satisfy 3x > 0 and 5x-2 > 0

So, x = 1.37 (cor. to 2 d.p.)

ln (3x) + ln (5x-2)=3
ln [3x(5x-2)]=ln (e^3)
ln (15x^2-6x)=ln (e^3)
15x^2-6x=e^3
15x^2-6x-e^3=0
x=[6士(36+60e^3)sqrt.]/30