Question 143986
How would I solve the below problem...THX tutors!!! :) 
{{{log(16,x) + log(8,x) + log(4,x) + log(2,x) = 6}}} 
<pre><font size = 4 color = "indigo"><b>
Use the 'change of base' formula:

{{{ log(GIVEN_BASE,(A)) = log(NEW_BASE,(A))/log(NEW_BASE,(GIVEN_BASE)) }}}

on each of the first three terms on the left, using {{{2}}} as the {{{NEW_BASE}}}

{{{log(2,x)/log(2,16) + log(2,x)/log(2,8) + log(2,x)/log(2,4) + log(2,x) = 6}}}
 
Now we can evaluate those three denominators by using the definition of
logarithm:

   {{{log(B,A)=C}}} is equivalent to {{{B^C=A}}}

For {{{log(2,16)}}}, we set it equal to the unknown "y":

{{{log(2,16)=y}}}

Then it is equivalent to {{{2^y=16}}} or {{{2^y=2^4}}}, so {{{y = 4}}},
therefore {{{log(2,16)=y=4}}}  

------

For {{{log(2,8)}}}, we set it equal to the unknown "z":

{{{log(2,8)=z}}}

Then it is equivalent to {{{2^z=8}}} or {{{2^z=2^3}}}, so {{{z = 3}}},
therefore {{{log(2,8)=z=4}}} 

------

For {{{log(2,4)}}}, we set it equal to the unknown "w":

{{{log(2,4)=w}}}

Then it is equivalent to {{{2^w=4}}} or {{{2^w=2^2}}}, so {{{w = 2}}},
therefore {{{log(2,4)=w=2}}} 

------

Substitute these for the first three terms of:

{{{log(2,x)/log(2,16) + log(2,x)/log(2,8) + log(2,x)/log(2,4) + log(2,x) = 6}}}

{{{log(2,x)/4 + log(2,x)/3 + log(2,x)/2 + log(2,x) = 6}}}

Clear of fractions by multiplying every term by LCD = 12

{{{12*log(2,x)/4 + 12*log(2,x)/3 + 12*log(2,x)/2 + 12*log(2,x) = 12*6}}}

{{{3log(2,x) + 4log(2,x) + 6log(2,x) + 12log(2,x) = 12*6}}}

{{{(3+4+6+12)log(2,x) = 72}}}

{{{25*log(2,x) = 72}}}

Divide both sides by 25

{{{25*log(2,x)/25 = 72/25}}}

{{{log(2,x) = 72/25}}}

Now use the definition of logarithm again to rewrite
that as

{{{drawing(50,50,0,2,-3,2, locate(0,0,x=2^(72/25)) )}}}

Find that on a TI calculator by typing 

2^(72/25)

then pressing ENTER

{{{x = 7.361501205}}}  

Edwin</pre>