Question

verify answer:

write this as a single log : 2ln x - 4ln (1/y) - 3ln (xy)

I get: ln (y/4x)

Answer 1

i think it's just ln(y/x)

Answer 2

@blurbendy How did you get ln(y/x) ? Thanks.

Answer 3

\[2lnx - 4 \ln \frac{ 1 }{ y } - 3\ln (xy) \]
\[lnx^2 -4 \left[ \ln1 - lny \right] -3\ln(xy)\]
since ln1 =0 you get:
\[lnx^2 +lny^4-lnx^3 -lny^3 \]
\[\ln \frac{ x^2y^4 }{ x^3y^3}\]

since you are dividing you subtract exponents in the x : 2- 3 = -1 therefore the x goes in the denominator and in the y : 4-3 =1 so you get

\[\ln \frac{ y }{ x }\]

Answer 4

@Directrix

Answer 5

@camilasanchez Correct
Answers without work are not helpful (my opinion).