SOLUTION: What is the solution to (log base 16 of x) + (log base 4 of x ) + (log base 2 of x) = 7

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Question 16747: What is the solution to
(log base 16 of x) + (log base 4 of x ) + (log base 2 of x) = 7
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
There may be an easier way, but I can't think of it. Let's change to ln using the formula log+%28b%2C+x%29+=+%28ln+x%29%2F%28ln+b%29+.

%28lnx%29%2F%28ln+16%29+%2B+%28ln+x%29%2F%28ln+4%29+%2B+%28ln+x%29%2F%28ln2%29+=+7

%28lnx%29%2F%28ln+2%5E4%29+%2B+%28ln+x%29%2F%28ln+2%5E2%29+%2B+%28ln+x%29%2F%28ln2%29+=+7

%28lnx%29%2F%284%2Aln+2%29+%2B+%28ln+x%29%2F%282%2Aln+2%29+%2B+%28ln+x%29%2F%28ln2%29+=+7

Multiply both sides by the LCD which is 4%2A+ln+2:


ln+x+%2B+2+%2A+ln+x+%2B+4+%2A+ln+x+=+28+%2Aln+2
7+%2A+ln+x+=+28%2A+ln+2

Divide both sides by 7:
ln+x+=+4+ln+2
ln+x+=+ln+2%5E4
ln+x+=+ln+16

By definition of equality of logarithms, or by raising both sides as a power of e,
x=+16

It checks: log+%2816%2C+16%29+%2B+log+%284%2C+16%29+%2B+log+%282%2C+16%29+=+1+%2B+2+%2B+4+=+7

R^2 at SCC