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Logarithmic equation

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365
1
+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 :)

Jul 20, 2014

#1
+95951
+10

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.......

Jul 20, 2014

#1
+95951
+10

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.......

CPhill Jul 20, 2014