+12 Hi. I'm having trouble with this logarithmic equation I'm being asked to do without a calculator. The answer is 8 - the only problem I have is getting there. It's the $${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{16}}}}$$ that's really messing me up. A little help?
Thanks, guys :)
+128460 Let's take this by parts
log32 + log 161/2 = lg 9 + log 4 = log (9*4) = log(36) .... and.....
[log 12 - log 2] = log (12/2) = log (6)
So, the first fraction becomes.....log(36)/log(6) = log(6)2 / log(6) = 2log(6)/log(6) = 2
And the numerator of the second fraction is.......
[2 - log 25 + 2log8] = ] 2 - log25 + log 82 ] = [2 - log25 + log64]
Notice, in terms of logs, that 2 can be written as log 100
So we have..
[log100- log25 + log 64] = [ log (100/25) + log64] = [log(4) + log (64)] =
log (4 * 64) = log(256) = log(16)2 = 2log(16)
And the denominator is....
log 3√16 = log 16 (1/3) = (1/3) log 16
And the second fraction becomes...
[2log(16)] / [(1/3)log(16)] = 2/(1/3) = 6
So...combining the result of the first fraction and the second fraction, we have
2 + 6 = 8
And there you go.......
+128460 Let's take this by parts
log32 + log 161/2 = lg 9 + log 4 = log (9*4) = log(36) .... and.....
[log 12 - log 2] = log (12/2) = log (6)
So, the first fraction becomes.....log(36)/log(6) = log(6)2 / log(6) = 2log(6)/log(6) = 2
And the numerator of the second fraction is.......
[2 - log 25 + 2log8] = ] 2 - log25 + log 82 ] = [2 - log25 + log64]
Notice, in terms of logs, that 2 can be written as log 100
So we have..
[log100- log25 + log 64] = [ log (100/25) + log64] = [log(4) + log (64)] =
log (4 * 64) = log(256) = log(16)2 = 2log(16)
And the denominator is....
log 3√16 = log 16 (1/3) = (1/3) log 16
And the second fraction becomes...
[2log(16)] / [(1/3)log(16)] = 2/(1/3) = 6
So...combining the result of the first fraction and the second fraction, we have
2 + 6 = 8
And there you go.......
