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Write 2log35 + log32 as a single logarithm.

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Using the log law a log(z) = log(z^a),
2 log(35) = log(35^2) = log(1225)

Using the log law log(a) + log(b) = log(a*b),
2log35 + log32 = log(1225*32) = log(39200)

You didn't state what base the logs have, so I assume it's 10. Then the answer in the requested form is log(10)(39200).
User Kasheena
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