Question

Yet another logarithm

Answer 1

Is 18 the base of the second logarithm or...?

Answer 2

Somehow i am having problem with latex , 18 is the base but the 8 is going to the power and 1 is going in the base

Answer 3

Question ?

Answer 4

{18}

Answer 5

For bases `\log_{18}`

Answer 6

18 is the base , Yes it is not coming properly it tried it @mathslover

Answer 7

18 IS THE BASE EVERYBODY

Answer 8

\(\huge \log_xlog_{18}(\sqrt{2}+\sqrt{18}) =\frac{ 1 }{ 3 }\)

Answer 9

Yes

Answer 10

\[\large \log_x \left(\log_{18}\left(\sqrt2 + \sqrt{18}\right)\right) = \dfrac{1}{3}\]\[x > 0, x \ne 1\]\[\large \log_{18}\sqrt{2} + 3\sqrt{2} = x^{1/3}\]\[\large \log_{18} 4\sqrt{2} = x^{1/3}\]\[\large \dfrac{2\log 2 + 0.5 \log 2 }{2\log 3 + \log2} = x^{1/3}\]\[\large \dfrac{5\log2}{4\log3 + 2\log2}=x^{1/3}\]\[\large \dfrac{\log(2^5)}{\log(3^4 \cdot 2^2)} = x^{1/3}\]\[\large \log_{3^4 \cdot 2^2}{2^5} = x^{1/3} \]Cube both sides...

Answer 11

\[\huge I got x= 128*\sqrt{2}\]

Answer 12

By my method

Answer 13

Note that \((3^4\cdot 2^2) = (3^2 \cdot 2 )^2 \)

Answer 14

\[\dfrac{1}{2 }\log_{18}32 = x^{1/3}\]

Answer 15

What's your method?

Answer 16

\[\large 18^{x^{1/3}} = 4\sqrt{2}\]\[\large x^{1/3} = \dfrac{\log_2 4\sqrt 2}{ \log_2 18} \]\[\large= \dfrac{5}{2 \log_2 {18}}\]So many ways to do the same darn question.