Question 1074217
<pre><b>
{{{2log(2,(x)) + log(2,(x-1)) = 1+log(2,(35x + 100))}}}

Use the rule of logarithms on the first term that allows
a coefficient of a logarithm to be moved to the exponent
of the expression which the logarithm is of.

{{{log(2,(x^2)) + log(2,(x-1)) = 1+log(2,(35x + 100))}}}

Since log<sub>2</sub>(2) = 1 we can replace the 1 on the
right by that:

{{{log(2,(x^2)) + log(2,(x-1)) = log(2,(2))+log(2,(35x + 100))}}}

Use the rule of logarithms that allows a sum of logarithms
to be written as the logarithm of the product of the 
expressions which the logarithms are of.

{{{log(2,(x^2(x-1))) = log(2,(2^""(35x + 100)))}}}

Use the rule of logarithms  that if {{{log(B,A)=log(B,C)}}}, then A=C

{{{x^2(x-1) = 2(35x + 100)}}}

{{{x^3-x^2 = 70x + 200}}}

{{{x^3-x^2 - 70x - 200=0}}}

Edwin</pre>