SOLUTION: I need to prove why the following equations are equal to each other. I have tried factoring, and using the rules for logarithms (meaning using multiplication when adding and divis

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Question 238822: I need to prove why the following equations are equal to each other. I have tried factoring, and using the rules for logarithms (meaning using multiplication when adding and division when subtracting). I still can not figure it out. I do not know how to use subscripts for math problems on a computer so if you don't understand what I am asking it is completely understandable. But for now I will use log(2) of 3, meaning a log with a base of two, of three. Here is the equation:
[6log(2)3] / [1 + log(2)3] = [6ln3] / [ln2 + ln3]
Thank you,
Heather
Found 2 solutions by stanbon, kensson:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
[6log(2)3] / [1 + log(2)3] = [6ln3] / [ln2 + ln3]
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Working with the left side:
6[ln(3)/ln(2)] / [1 + ln(3)/ln(2)]
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= [6ln(3)]/[ln(2)] divided by [ln(2) + ln(3)]/[ln(2)]
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If you invert the denominator and multiply the ln(2)'s cancel
and you end up with:
= [6ln(3)] divided by [ln(2) + ln(3)]
which is the right side you your original equation.
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Comment: The key here is that you see that log(2)3 = ln(3)/ln(2)
That is using the Change of base Rule.
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Cheers,
Stan H.

Answer by kensson(21) (Show Source):
You can put this solution on YOUR website!
Using the change of base rule, I reckon the LHS works out to be:

If we multiply top and bottom by ln 2, we get:

as required.