SOLUTION: Write ({{{ log( 3, 7 ) * log( 4, 3 ) * log( 7, 6 ) }}}) as a single logarithm.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Write ({{{ log( 3, 7 ) * log( 4, 3 ) * log( 7, 6 ) }}}) as a single logarithm.      Log On


   



Question 935990: Write (+log%28+3%2C+7+%29+%2A+log%28+4%2C+3+%29+%2A+log%28+7%2C+6+%29+) as a single logarithm.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
this was a doozy, but i think i was able to figure it out.

by the log conversion formula, you get:

log%283%2C7%29+=+log%287%29%2Flog%283%29 and log%284%2C3%29+=+log%283%29%2Flog%284%29 and log%286%2C7%29+=+log%286%29%2Flog%287%29

you get the following equation:

(+log%28+3%2C+7+%29+%2A+log%28+4%2C+3+%29+%2A+log%28+7%2C+6+%29+) = log%287%29%2Flog%283%29+%2A+log%283%29%2Flog%284%29+%2A+log%286%29%2Flog%287%29

log%287%29 in the numerator and denominator cancel out, and log%283%29 in the numerator and denominator cancel out and you are left with log%286%29+%2F+log%284%29


by the log base conversion formula, you get:

log%284%2C6%29+=+log%286%29+%2F+log%284%29 and so your final solution is:

(+log%28+3%2C+7+%29+%2A+log%28+4%2C+3+%29+%2A+log%28+7%2C+6+%29+ = +log%284%2C6%29.

to confirm the solution is correct, you have to go back to the log conversion formulas.

+log%284%2C6%29.= log%286%29%2Flog%284%29 = 1.29248125.

(+log%28+3%2C+7+%29+%2A+log%28+4%2C+3+%29+%2A+log%28+7%2C+6+%29+) = log%287%29%2Flog%283%29+%2A+log%283%29%2Flog%284%29+%2A+log%286%29%2Flog%287%29 = 1.29248125.

they give you the same value, so they're equivalent.

your solution is that the expression becomes log%284%2C6%29