+0  
 
0
347
3
avatar

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

how to explain this?

Guest Jan 6, 2015

Best Answer 

 #2
avatar+87301 
+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

 

 

CPhill  Jan 7, 2015
 #1
avatar+92781 
+5

Mmm interesting.

Let 

 

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

Melody  Jan 7, 2015
 #2
avatar+87301 
+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+92781 
0

Thanks Chris 

It is always good to get a couple of different takes  

Melody  Jan 7, 2015

10 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.