SOLUTION: solve for x. log(2)^x=8+9log(x)^2

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Question 408507: solve for x. log(2)^x=8+9log(x)^2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
It is very difficult to understand what your equation is. Is it
+log%28%282%5Ex%29%29=8%2B9log%28%28x%5E2%29%29+
or
+%28log%28%282%29%29%29%5Ex+=+8+%2B+9%28log%28%28x%29%29%29%5E2
or
+log%282%2C+%28x%29%29+=+8+%2B+9log%28x%2C+%282%29%29+
or something else?

Tutors are more likely to help when the problem is clearly stated. Logarithmic expressions can be particularly difficult. With logarithms I suggest you learn how to read them and then post them in part English and part Math terms. For example,
log%283%2C+%28x%2By%29%29
is the "base 3 log of (x+y)". Note the use of parentheses. The "base 3 log of x + y" would be:
log%283%2C+%28x%29%29+%2B+y
which is not the same thing.

BTW, if your equation was
+log%282%2C+%28x%29%29+=+8+%2B+9log%28x%2C+%282%29%29+
(or something close to it), then there are no exponents in this equation!. This equation is "the base 2 log of x = 8 + 9 times the base x log of 2".