Question

no

Answer 1

Cross multiply maybe?

Answer 2

2 hints:

a) \(\Large \frac{x^a}{x^b} = x^{a-b}\)

b) see rule 2 on the link below

http://www.purplemath.com/modules/logrules.htm

Answer 3

good

Answer 4

do you see how to isolate x from there?

Answer 5

yep, log(a/b) is just a constant

Answer 6

so basically raise both sides to the 1/[log(a/b)] power

Answer 7

it's a bit ugly but it's still doable

Answer 8

correct

\[\LARGE x = \left(\frac{b}{a}\right)^{\frac{1}{\log\left(\frac{a}{b}\right)}}\]

and that's as far as you can go in terms of simplifying it

Answer 9

well I guess you could do this trick
\[\LARGE x = \left(\frac{b}{a}\right)^{\frac{1}{\log\left(\frac{a}{b}\right)}}\]
\[\LARGE x = \left(\left(\frac{a}{b}\right)^{-1}\right)^{\frac{1}{\log\left(\frac{a}{b}\right)}}\]
\[\LARGE x = \left(q\right)^{-\frac{1}{\log\left(q\right)}} \ ... \ \text{Let } q = \frac{a}{b}\]
but that's all I can think of in terms of simplification (it's not much of a simplification really)

Answer 10

you're welcome

Answer 11

how exactly?

Answer 12

Oh I see, we can use the change of base formula
\[\LARGE x = \left(q\right)^{-\frac{1}{\log\left(q\right)}}\]
\[\LARGE x = \left(q\right)^{-\frac{1}{\log_{10}\left(q\right)}}\]
\[\LARGE x = \left(q\right)^{-\frac{1}{\log\left(q\right)/\log(10)}}\]
\[\LARGE x = \left(q\right)^{-\frac{\log(10)}{\log\left(q\right)}}\]
\[\LARGE x = \left(q\right)^{-\log_q(10)}\]
\[\LARGE x = \left(q\right)^{\log_q(10^{-1})}\]
\[\LARGE x = 10^{-1}\]
\[\LARGE x = \frac{1}{10}\]

Answer 13

So you can have the shortcut rule

\[\LARGE \frac{1}{\log_{x}(y)} = \log_{y}(x)\]

Answer 14

you're welcome

Answer 15

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