Logarithm help (express as a single log)

2loga x3 - log x2 - 3 log x

can you please explain to me how to solve this equation?
thank you so much <3

Assume all logs to same base a.
log x^3 =3logx
logx^2 = 2logx

So 2logx^3 - logx^2 -3logx = 6logx - 2logx - 3logx = logx

Hope this helps.

Are they all log base a?? And do you mean:

x3 = x³, x2 = x²

...if so, then just use the addition/subtraction rules:

log(a) + log(b) = log(a * b) and log(a) - log(b) = log(a / b)

furthermore: nlog(a) = log(a^n)
-->

2log_a(x³) = log_a((x³)²) = log_a(x⁶)

3log_a(x) = log_a(x³)
-->

2log_a(x³) - log_a(x²) - 3log_a(x)
-->

log_a(x⁶) - log_a(x²) - log_a(x³)
-->

log_a(x⁶ / (x² * x³)) = log_a(x⁶ / x⁵) = log_a(x)

trick: logx + logy = log(xy)
and, logx - logy = log(x/y)

2loga x3 - log x2 - 3 log x
log(a2 x6) - log(x2) - log(x3)
log(a^2 c^6) - log(x6)
log(a2)
2*log(a)

2log_a x^3 - log_a x^2 - 3 log_a x =
log_a (x^3)^2 - log_a x^2 - log_a x^3 =
log_a (x^3)^2 / x^2*x^3 =
log_a (x^6 / x^5) =
log_a x