SOLUTION: Hi goodnight, can you solve for x: log 2X^2 + logX = log32 - logX I did the question already just not sure of the answer please help

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Question 909304: Hi goodnight, can you solve for x:
log 2X^2 + logX = log32 - logX
I did the question already just not sure of the answer please help
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
log+%282X%5E2%29+%2B+log%28X%29+=+log%2832%29+-+log%28X%29
log+%282X%5E2%29+%2B+log%28X%29+%2Blog%28X%29=+log%2832%29
log+%282X%5E2%29+%2B+2log%28X%29+=+log%2832%29+
log+%282X%5E2%29+%2B+log%28X%5E2%29+=+log%2832%29+
log+%28%282X%5E2%2AX%5E2%29%29+=+log%2832%29
log+%282X%5E4%29+=+log%2832%29+ if log equal,then
2X%5E4+=+32+
X%5E4+=+32%2F2+
X%5E4+=+16
X+=root%284%2C+16%29
X+=root%284%2C+2%5E4%29
X+=2