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If log(2) x xx log(3) x = log(2) x + ...

If ` log_(2) x xx log_(3) x = log_(2) x + log_(3) x`, then find x .

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The correct Answer is:
x = 1, 6
` log_(2) x xx log_(3) x = log_(2) x + log_(3) x`
` rArr (log x)/(log 2) *(log x)/(log 3) =(log x)/(log 2) +(log x)/(log 3) `
` rArr log x = 0`
` or (log x)/(log 2*log 3) = 1/(log 2) + 1/(log 3) `
` rArr x = 1 or log x = log 2+ log 3`
`rArr x = 1 or log x = log 6`
` rArr x = 1 or x = 6`
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