Question 279629: prove that : log 3 of base 2 multiply log 2 of base 3 = 1
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! log(2,3) * log(3,2) = 1
by the conversion of logs from one base to another formula, we get:
log(2,3) = log(10,3) / log(10,2) and we get:
log(3,2) = log(10,2) / log(10,3)
substituting these equivalencies into the original equation, we get:
log(10,3) / log(10,2) * log(10,2) / log(10,3) = 1
this is equivalent to:
(log(10,3) * log(10,2)) / ((log(10,2) * log(10,3)) = 1
log(10,3) in the numerator and denominator cancel out.
log(10,2) in the numerator and denominator cancel out.
we are left with 1 = 1 which is true confirming that the identity is correct.
we can solve this another way as well.
original equation is:
log(2,3) * log(3,2) = 1
log(2,3) = y if an only if 2^y = 3
log (3,2) = z if and only if 3^z = 2
take the log of both sides of the equation of 2^y = 3 and you get:
log(2^y) = log(3)
by the rules of logarithms, this becomes:
y*log(2) = log(3)
divide both sides of this equation by log(2) to get:
y = log(3)/log(2)
use your calculator to solve for y to get y = 1.584962501 **********
now take the log of both sides of the equation of 3^z = 2 and you get:
log(3^z) = log(2)
by the rules of logarithms, this becomes:
z*log(3) = log(2)
divide both sides of this equation by log(3) to get:
z = log(2)/log(3)
use your calculator to get z = .630929754
your original equation is:
log(2,3) * log(3,2) = 1
we set log(2,3) = y and we set log(3,2) = z to get:
y * z = 1
we solved for y to get y = 1.584962501 and we solved for x to get z = .630929754.
we substitute these values for y and z in the equation to get:
1.584962501 * .630929754 = 1
we simplify to get 1 = 1 confirming that the original equation is true.
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