SOLUTION: Let (((log(a,X))))=C and (((log(b,X))))=D. Find the general statement that expresses (((log(ab,X))))in term of C and D. I already know the answer is (CD)/(C+D) I cannot figur

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Let (((log(a,X))))=C and (((log(b,X))))=D. Find the general statement that expresses (((log(ab,X))))in term of C and D. I already know the answer is (CD)/(C+D) I cannot figur      Log On


   

Question 222790: Let (((log(a,X))))=C and (((log(b,X))))=D. Find the general statement that expresses (((log(ab,X))))in term of C and D.
I already know the answer is (CD)/(C+D)
I cannot figure out how to write a proof for this question.
I know all of the logarithmic principles as well but I can only come up with either (C)(D) as my answer or (C+D). I can't get a fraction.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I found that it helps to work backwards for a bit to find out how to go forward. In this case, I broke down log%28ab%2C%28x%29%29 to get:

log%28ab%2C%28x%29%29


log%2810%2C%28x%29%29%2Flog%2810%2C%28ab%29%29 (used the change of base formula here)


log%2810%2C%28x%29%29%2F%28log%2810%2C%28a%29%29%2Blog%2810%2C%28b%29%29%29 (used the identity log%28b%2C%28A%2AB%29%29=log%28b%2C%28A%29%29%2Blog%28b%2C%28B%29%29)


Then I realized that I needed to find substitutions for log%2810%2C%28a%29%29 and log%2810%2C%28b%29%29 (to somehow get 'x' in there). So I figured I'd solve for those respective expressions like so:



log%28a%2C%28x%29%29=C Start with the first equation.


log%2810%2C%28x%29%29%2Flog%2810%2C%28a%29%29=C Use the change of base formula.


log%2810%2C%28x%29%29=C%2Alog%2810%2C%28a%29%29 Multiply both sides by log%2810%2C%28a%29%29.


log%2810%2C%28x%29%29%2FC=log%2810%2C%28a%29%29 Divide both sides by C.


So this means log%2810%2C%28a%29%29=log%2810%2C%28x%29%29%2FC


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log%28b%2C%28x%29%29=D Start with the second equation.


log%2810%2C%28x%29%29%2Flog%2810%2C%28b%29%29=D Use the change of base formula.


log%2810%2C%28x%29%29=D%2Alog%2810%2C%28b%29%29 Multiply both sides by log%2810%2C%28b%29%29


log%2810%2C%28x%29%29%2FD=log%2810%2C%28b%29%29 Divide both sides by D.


So this means log%2810%2C%28b%29%29=log%2810%2C%28x%29%29%2FD

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Now let's get back to the main problem:


log%28ab%2C%28x%29%29 Start with the given expression.


log%2810%2C%28x%29%29%2Flog%2810%2C%28ab%29%29 Use the change of base formula.


log%2810%2C%28x%29%29%2F%28log%2810%2C%28a%29%29%2Blog%2810%2C%28b%29%29%29 Use the identity log%28b%2C%28A%2AB%29%29=log%28b%2C%28A%29%29%2Blog%28b%2C%28B%29%29 to break up the denominator.


Now here's where the substitutions come into play:


log%2810%2C%28x%29%29%2F%28log%2810%2C%28x%29%29%2FC%2Blog%2810%2C%28x%29%29%2FD%29 Plug in log%2810%2C%28a%29%29=log%2810%2C%28x%29%29%2FC and log%2810%2C%28b%29%29=log%2810%2C%28x%29%29%2FD


From here, it's just boils down to algebraically simplifying the expression:


Multiply the first inner fraction by D%2FD.


Multiply the second inner fraction by C%2FC.


Combine the lower fractions.


Multiply log%2810%2C%28x%29%29 by the reciprocal of the lower fraction.


Multiply


%28CD%2Alog%2810%2C%28x%29%29%29%2F%28log%2810%2C%28x%29%29%28D%2BC%29%29 Factor out the GCF log%2810%2C%28x%29%29 from the denominator.


Cancel out the common terms.


%28CD%29%2F%28D%2BC%29 Simplify


%28CD%29%2F%28C%2BD%29 Rearrange the terms (trivial, but I figured that I'd match the book).


So this means log%28ab%2C%28x%29%29=%28CD%29%2F%28C%2BD%29 where log%28a%2C%28x%29%29=C and log%28b%2C%28x%29%29=D