Please solve: 2logx = log32 + log2

please explain as well.

Two rules of logarithms are important here:

1) Log(A) + Log(B) = Log(A*B)

2) A*Log(B) = Log(B^A)

So, we can change the left hand side to:

2log(x) = log(x^2)

The right hand side becomes:

log(32) + log(2) = log(32*2) = log(64)

So we now have:

log(x^2) = log(64)

x^2 = 64

Initially, we have x = 8 or -8. However, -8 is not a possible answer because we cannot take the logarithm of a negative. When you look at the initial equation, x = -8 would result in taking the log of a negative number.

Thus, the only answer is x = 8

a log(x) = log (x^a)
log(a) + log(b) = log (a * b)
log(a) = log(b) ==> a = b

basic rules...

so 2 log(x) = log(32) + log(2)
==> log(x^2) = log(32 * 2) = log(64)
==> x^2 = 64
==> x = +/- 8

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This is the same question you have asked and the people pretty much explain it as well as give you the answer on it :)

2logx = log(32*2) = log64.............. log(x*x) = log64............x*x = 64........x = +8, -8