SOLUTION: (square root of 7)^x+4=49x ln2x+ln8x=ln17

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Question 42666: (square root of 7)^x+4=49x

ln2x+ln8x=ln17
Answer by astromathman(21) About Me  (Show Source):
You can put this solution on YOUR website!
For the first problem, do you mean sqrt%287%29%5E%28x%2B4%29=49%5Ex?
If so it can be solved easily. As written it is ambiguous and other ways to interpret it would make it harder to solve.
Note that sqrt%287%29+=+7%5E%281%2F2%29 and 49+=+7%5E2. We could rewrite the problem as 7%5E%28%28x%2B4%29%2F2%29=7%5E%282x%29, so %28x%2B4%29%2F2=2x, leading to x+=+4%2F3
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Second problem:
ln+2x+%2B+ln+8x+=+ln+17
ln+2+%2Bln+x+%2B+ln+8+%2B+ln+x+=+ln+17
2%2Aln+x+=+ln+17+-+ln+2+-+ln+8
ln+x%5E2+=+ln+%2817%2F%282%2A8%29%29+
x%5E2+=+17%2F16
x+=+sqrt%2817%29%2F4