Logarithm help please!

I've been trying to do this question and I have no idea how to even begin. It makes no sense to me:

Given that loga(x^2) = 3 show without the use of a calculator that

loga(x^2) + ln(x^3)+6loga(x) = 17/3ln(x)

can someone please help?!

thanks!

so I've changed it to 2loga(x) + 3lnx + 6loga(x) + (17/3)lnx

what base do I need to change? loga?

You should not have the (17/3)lnx

That is what you are trying to get

So the rest needs to be in ln

ok so I got 17lnx by doing this:

2loga(x) + 3lnx + 6loga(x)

2(lnx/lna) + 3lnx + 6(lnx/lna)

I multiplied it by lna to get 2lnx + lna3lnx +6lnx

2lnx + 9lnx + 6lnx = 17lnx

I don't know where I went wrong, i'm supposed to get 17/3lnx

You have 3lnalnx and then you change that to 9lna ... Why

From the first line

2logx + 3lnx + 6logx

You can use the fact that logx = 1.5



Do you have a picture of the question as it does not seem to work

Am pretty sure the question as stated is not true. I did assume logax was $\log_ax$ though.

I've been trying to do this question and I have no idea how to even begin. It makes no sense to me:

Given that loga(x^2) = 3 show without the use of a calculator that

loga(x^2) + ln(x^3)+6loga(x) = (17/3)ln(x)

can someone please help?!