How do you solve log(y) = 1/4 log16 + 1/2 log49log(y)=14log16+12log49?

1 Answer
Aug 2, 2015

log y = 1/4 log 16 + 1/2 log 49 = log 2 + log 7 = log 14logy=14log16+12log49=log2+log7=log14

So y = 14y=14

Explanation:

If a > 0a>0 then log a^b = b log alogab=bloga

If a, b > 0a,b>0 then log ab = log a + log blogab=loga+logb

So:

log y = 1/4 log 16 + 1/2 log 49 = 1/4 log 2^4 + 1/2 log 7^2logy=14log16+12log49=14log24+12log72

= 1/4*4 log 2 + 1/2 * 2 log 7 = log 2 + log 7 = log (2*7)=144log2+122log7=log2+log7=log(27)

= log 14=log14

That is: log y = log 14logy=log14

If log a = log bloga=logb then a = ba=b

So y = 14y=14