SOLUTION: please help me solve: {{{ 2 * log(4,(x)) + 3 * log(8,(x)) = 10}}} thanks

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: please help me solve: {{{ 2 * log(4,(x)) + 3 * log(8,(x)) = 10}}} thanks      Log On


   



Question 332329: please help me solve:
+2+%2A+log%284%2C%28x%29%29+%2B+3+%2A+log%288%2C%28x%29%29+=+10
thanks

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Logarithmic equations are usually easier to solve when the bases of all the logarithms are the same. This allows one to use the properties of logarithms. So the simplest solution to this equation probably starts with changing the bases of the logarithms, using the change-of-base formula. 4 and 8 are not (obvious) powers of each other but they are both powers of 2 so we change both logarithms in base 2 logarithms:


Since 2%5E2+=+4 then log%282%2C+%284%29%29+=+2. Similarly, log%282%2C+%288%29%29+=+3. Substituting we get:
+2+%2A+log%282%2C%28x%29%29%2F2+%2B+3+%2A+log%282%2C%28x%29%29%2F3+=+10
which simplifies to:
+log%282%2C%28x%29%29+%2B+log%282%2C%28x%29%29+=+10
Now that the bases are the same we can use a property of logarithms to combine the remaining logs:
log%282%2C+%28x%2Ax%29%29+=+10
or
log%282%2C+%28x%5E2%29%29+=+10
Now we can write this in exponential form:
2%5E10+=+x%5E2
Solving this equation we get:
2%5E5+=+x or -2%5E5+=+x
which simplifies to:
32+=+x or -32+=+x

When solving logarithmic equations you must check your answers. You must reject any "solutions" which would cause the argument of a logarithm to become negative. We can easily see that the arguments of both logarithms of
+2+%2A+log%284%2C%28x%29%29+%2B+3+%2A+log%288%2C%28x%29%29+=+10
will become negative if -32+=+x. So we must reject this "solution". (If only one argument becomes negative we would still reject the solution.)

So the only solution to your equation is:
32+=+x