How do you condense $3 log y 3 + log y 2$ $3 logy3 + logy 2$ ?

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Answer 1 · 2018-05-24T09:10:04

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I think you wanted to write $3 {log y}^{3} + {log y}^{2}$ $3logy^3+logy^2$

In that case, by logarithm rules

$3 \cdot 3 log y + 2 log y = 9 log y + 2 log y = 11 log y = {log y}^{11}$ $3·3logy+2logy=9logy+2logy=11logy=logy^11$

If you write, exactly that you want, then, by logarithm rules

$3 log y + log 3 + log y + log 2 = 4 log y + log 3 + log 2$ $3logy+log3+logy+log2=4logy+log3+log2$

The rules we have used are

$log x \cdot y = log x + log y$ $logx·y=logx+logy$
${log x}^{y} = y log x$ $logx^y=ylogx$

How do you condense 3logy3+logy23 logy3 + logy 2?