Question

Can someone help me with this log problem?

Answer 1

\[5.4(x) ^{2.8}-3.1=12.9\]

Answer 2

Well, this'd be murder without a calculator, so I take it you have one at your disposal?

Answer 3

yep : )

Answer 4

Good. For now, treat that nasty \(\large (x)^{2.8}\) as a single unknown, and solve for it, just like you would any typical variable. Hop to it ^_^

Answer 5

Yeah I got \[x^{2.8}=2.9629\] what do i do next?

Answer 6

Logs aren't strictly necessary here, especially since you have a calculator, but if logs MUST be used, then take the natural log (ln) of both sides.

Answer 7

ok so what does that mean? ln\[lnx ^{2.8}=\ln2.9629\] I still don't know how to solve hat

Answer 8

Relax.
Let me rewrite that for you.
\[\Large \ln(x^{2.8}) = \ln(2.9629)\]
key in the right-hand side on your calculator.

Answer 9

That is to say... what IS ln(2.9629) ?

Answer 10

ohhh ok so that's 1.086

Answer 11

I'm not using a calculator here, so I'm going to take your word for it, ok? \[\Large \ln (x^{2.8}) = 1.086\]
Good.

Now,there's a certain property of logs... roughly speaking... it turns exponents into factors... THIS property
\[\Large \ln (x^p) = p \ \ln(x)\]
I believe this applies to your current equation here, quite conveniently...

Answer 12

so than \[2.8lnx=\ln2.962\]
\[\ln x=\ln2.29629/2.8\]
How do you get x by itself

Answer 13

Don't use ln(2.962) you already established that that's equal to 1.086, remember? :)

Answer 14

\[\Large \ln(x) = \frac{1.086}{2.8}\]
Should do it.
Simplify first...

Answer 15

ok than do you e^ and find x?

Answer 16

That is correct ^_^

Answer 17

thanks!!!

Answer 18

No problem :)

Answer 19

nice walkthrough there Terenzreignz

Answer 20

Thanks :)
Please call me TJ ^^