Question 1057131
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I just need someone to explain this and why 2 is the answer:
3^(log base 9 (4))= 2

I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.
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The problem is solved in two steps:


<pre>
1.  {{{log(9,(4))}}} = {{{log(3,(2))}}}


    Indeed, if {{{log(3,(2))}}} = x, it means (by the definition of logarithm) that {{{3^x}}} = 2.
    Then  {{{(3^x)^2}}} = {{{2^2}}} = 4, which implies  {{{3^(2x)}}} = 4, which is the same as {{{(3^2)^x}}} = 4,  or  {{{9^x}}} = 4.
    The last equality means that {{{log(9,(4))}}} = x.  Hence,  {{{log(9,(4))}}} = {{{log(3,(2))}}},  QED.



2.  {{{3^(log(3,(2)))}}} = 2.    (<---- {{{b^(log(b,(x)))}}} = x  for any x > 0 and for any b > 0,  b=/= 1.  
                                 This is the basic property of the logarithm, equivalent to the definition of the logarithm)
</pre>

On logarithms, see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/logarithm/what-is-the-logarithm.lesson>WHAT IS the logarithm</A>  

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/logarithm/Properties-of-the-logarithm.lesson>Properties of the logarithm</A>  

in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Logarithms</U>".