SOLUTION: How do I solve the logarithmic equation symbolically, log x^8=7+2 log x, we did not go over this in class? v/r Dean

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Question 21612: How do I solve the logarithmic equation symbolically, log x^8=7+2 log x, we did not go over this in class?
v/r Dean
Answer by NaiXLG(18) About Me  (Show Source):
You can put this solution on YOUR website!
Ahhhh yes, logarithims, gotta love 'em...


So, you begin by writing the original question
Solve for x:
log+x%5E8+=+7+%2B+2log+x

So, begin by rewriting the right logarithim as log+x%5E2 because the properties of logarithims states that a+log+b+=+log+b%5Ea+

So now you should have,
log+x%5E8+=+7+%2B+log+x%5E2
Bring the right logarithim to the left, so now we have all logs on the left, and all non-logs (constants) on the right.
log+x%5E8+-+log+x%5E2+=+7
Now using yet another property of logarithims, go ahead and simplify that left side into one single log....
Remember log+a+-+log+b+=+log%28a%2Fb%29
So, you'll have;
log%28x%5E8%2Fx%5E2%29+=+7
Now, use your properties of exponents to simplify that log a little more...
Remember x%5Em%2Fx%5En+=+x%5E%28m-n%29
So,
log%28x%5E6%29+=+7


Now, because the log has a base of ten, raise both sides of the equation to the tenth power....





10%5Elog%28x%5E6%29+=+10%5E7
The log cancels out, and you're left with
x%5E6+=+10%5E7
Now simply take the 6th root of x to make it just a single x, and no more, and take the 6th root of 10%5E7 as well. (What you do to one side of the equation, you must do to the other.)

x+=+10%5E%287%2F6%29
There you go....!