Calculate log(base12) (18) if log(base3)(2)=a?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to write `log_12 18` in terms of a, hence, you may start converting the argument 18 into a product, such that:

`log_12 18 = log_12 2*3^2`

Converting the logarithm of product into a sum of logarithms yields:

`log_12 18 = log_12 2 + log_12 3^2`

You need to write `log_12 2` in term of a, hence, you need to convert the base 12 into the base 3, such that:

`log_12 2 = (log_3 2)/(log_3 12)`

You need to write `12` as a product, such that:

`log_12 2 = (log_3 2)/(log_3 (3*4))`

Converting the logarithm of product into a sum of logarithms yields:

`log_12 2 = (log_3 2)/(log_3 3 + log_3 (2^2))`

Using the power property of logarithm yields:

`log_12 2 = (log_3 2)/(1 + 2log_3 2)`

Since the problem provides the information that `log_3 2 = a` , yields:

log_12 2 = a/(1 + 2a)

You need to write `log_12 3^2` in term of `a` , hence, you need to convert the base 12 into the base 3, such that:

`log_12 3 = (log_3 3)/(log_3 12)`

`log_12 3 = 1/(1 + 2a) =>log_12 3^2 = 2/(1 + 2a)`

You may write `log_12 18` in terms of `a` only, such that:

`log_12 18 = a/(1 + 2a) + 2/(1 + 2a)`

`log_12 18 = (a + 2)/(1 + 2a)`

Hence, evaluating log_12 18 in terms of a, yields `log_12 18 = (a + 2)/(1 + 2a).`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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It is given that `log_3 2 = a`

Take a common base x, and express `log_3 2` in terms of logarithm base x

`log_3 2 = (log_x 2)/(log_x 3) = a`

Now we have to determine `log_12 18`

Again write this in terms of base x

`log_12 18 = (log_x 18)/(log_x 12)`

= `(log_x 9*2)/(log_x 4*3)`

= `(log_x 2 + log_x 9)/(log_x 4 + log_x 3)`

= `(log_x 2 + 2*log_x 3)/(2*log_x 2 + log_x 3)`

Now `log_x 2 = a*(log_3 x)`

= `(a*(log_3 x) + 2*log_x 3)/(2*a*(log_3 x) + log_x 3)`

= `(a + 2)/(2*a + 1)`

The value of `log_12 18 = (a + 2)/(2*a + 1)`

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