If log12 (27) = a, then determine log6 (16)

[wind junco]

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[wind junco]

is it

$\log_{12}(27)=a$ and $\log_6(16)$?

[fresh tinsel]

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[wind junco]

Yea

[wind junco]

so apply logarithematic properties

$\log_b a = \frac{\log_x a}{\log_x b}$

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[wind junco]

Yeah, I couldnt seem to get the result

Even tho I got something

It was pretty ugly

[wind junco]

what you got?

let me see

[wind junco]

Cant remember

It was a fraction

I wrote it yesterday

$\frac83 \log_6(2) \log_3 (12) a$

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[wind junco]

is this what you got?

I dont think so

or else

I got something else

[wind junco]

Could you elaborate this

like this?

[wind junco]

alright

perhaps

I cant remember

this is a we got

$\log_{12}(27)=\log_{12}(3^3)=3\log_{12}(3)=3\frac{\ln(3)}{\ln(12)}$=a

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[wind junco]

is this fine?

$\log_6(16)=\log_6(2^3)=3\log_6(2)=3\frac{\ln(2)}{\ln(6)}$

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[wind junco]

take this as b

$3\frac{\ln(2)}{\ln(6)}=b$

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[wind junco]

now

$\frac{a}{b}=\frac{3\frac{\ln(3)}{\ln(12)}}{3\frac{\ln(2)}{\ln(6)}}$

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[wind junco]

now by cross multiplication,

$b=\frac{\ln(2)}{\ln(6)} \frac{\ln(12)}{\ln(3)}a$

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[wind junco]

now

$b=\log_6(2) \log_3(12) a$

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[wind junco]

@wind junco is this clear?

I wanted you to show your steps because I can asses you but fine

[wind junco]

Yep, thanks man 👍

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[wind junco]

if you do come through your notes send me/go through what you've done