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$${{log}}_{{\mathtt{0.5}}}{\left({\mathtt{19}}\right)}{\mathtt{\,\small\textbf+\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{19}}\right)} = {\mathtt{0}}$$

how to explain this?

 Jan 6, 2015

Best Answer 

 #2
avatar+94545 
+10

Here's a slightly different approach to Melody's

Note that we can write .5  as 1/2  = 2-1 

So we have

log2(19)  + log0.5 (19)   .....and...using the"change of base" rule, we can write

log(19) / log(2)  +  log(19) / log(.05) =

log(19) / log(2)  + log(19) / log(1/2) =

log(19) / log(2)  +  log(19) / log2-1 =

log(19) / log(2)  +  log(19) / (-1) log2 =

log(19) / log(2)  +  [ (-1) log(19) /  log2 ] = 

log(19) / log(2)  -  log(19) /  log2 =   0

 

 

 Jan 7, 2015
 #1
avatar+95334 
+5

Mmm interesting.

Let 

 

$$\\x=log_{0.5}(19)\qquad and \qquad y=log_2(19)\\
so\\
19=0.5^x\qquad\qquad and \qquad 19=2^y\\
19=(\frac{1}{2})^x\\
2^x=\frac{1}{19}\\
so\\
2^x\times 2^y=\frac{1}{19}\times 19\\
2^x\times 2^y=1\\
2^x\times 2^y=2^0\\
2^{x+y}=2^0\\
$therefore$\\
x+y=0\\
so\\
log_{0.5}(19)+log_2(19)=0$$

.
 Jan 7, 2015
 #2
avatar+94545 
+10
Best Answer

Here's a slightly different approach to Melody's

Note that we can write .5  as 1/2  = 2-1 

So we have

log2(19)  + log0.5 (19)   .....and...using the"change of base" rule, we can write

log(19) / log(2)  +  log(19) / log(.05) =

log(19) / log(2)  + log(19) / log(1/2) =

log(19) / log(2)  +  log(19) / log2-1 =

log(19) / log(2)  +  log(19) / (-1) log2 =

log(19) / log(2)  +  [ (-1) log(19) /  log2 ] = 

log(19) / log(2)  -  log(19) /  log2 =   0

 

 

CPhill Jan 7, 2015
 #3
avatar+95334 
0

Thanks Chris 

It is always good to get a couple of different takes  

 Jan 7, 2015

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