SOLUTION: 3logx+5 logx-6logx=64 what is x

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Question 1139059: 3logx+5 logx-6logx=64 what is x

Found 2 solutions by Theo, Alan3354:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
your equation is:

3 * log(x) + 5 * log(x) - 6 * log(x) = 64.

3 * log(x) is equal to log(x^3)
5 * log(x) is equal to log(x^5)
6 * log(x) is equal to log(x^6)

the equation becomes:

log(x^3) + log(x^5) - log(x^6) = 64

log(x^3) + log(x^5) is equal to log(x^3 * x^5) which is equal to log(x^8)

log(x^8) - log(x^6) is equal to log(x^8 / x^6) which is equal to log(x^2)

your equation becomes log(x^2) = 64.

this is true if and only if 10^64 = x^2.

take the square root of both sides of this equation to find that x = 10^32.

when x = 10^32, your original equation of 3 * log(x) + 5 * log(x) - 6 * log(x) = 64 becomes:

3 * log(10^32) + 5 * log(10^32) - 6 * log(10^32) = 64.

3 * log(10^32) is equal to 3 * 32 * log(10).
5 * log(10^32) is equal to 5 * 32 * log(10).
6 * log(10^32) is equal to 6 * 32 * log(10).

log(10) is equal to 1.

your equation therefore becomes:

3 * 32 * 1 + 5 * 32 * 1 - 6 * 32 * 1 = 64

factor out the 32 * 1 and you get:

32 * 1 * (3 + 5 - 6) = 64

simplify to get 32 * 1 * 2 = 64

simplify further to get 64 = 64, confirming the solution is correct.

you could also have used your calculator to get 3 * log(10^32) + 5 * log(10^32) - 6 * log(10^32) = 64.

here's a reference on logs you might find helpful.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm

any questions, email dtheophilis@gmail.com

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
3logx+5 logx-6logx=64 what is x
--------
2log(x) = 64
log(x) = 32
x = 10^32