Question 148642
{{{log(6,(b^2 + 2)) + log(6,(2)) = 2 }}} 
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Use this principle of logs on the left:{{{log(N,(X)) + log(N,(Y)) = log(N,(X*Y))}}}

{{{log(6,((b^2 + 2)*2)) = 2 }}}

{{{log(6,(2*(b^2 + 2))) = 2 }}}

{{{log(6,(2b^2 + 4)) = 2 }}}

Now use this principle of logs on that equation:

The log equation {{{log(N,X)=Y}}} can be rewritten 
as the exponential equation {{{X=N^Y}}}

{{{2b^2+4 = 6^2}}}

{{{2b^2+4 = 36}}}

Subtract {{{4}}} from both sides:

{{{2b^2=32}}}

Divide both sides by 2

{{{(2b^2)/2=32/2}}}

{{{b^2=16}}}

Subtract 16 from both sides:

{{{b^2-16=0}}}

Factor the left side:

{{{(b-4)(b+4)=0}}}

Set each factor = 0

{{{b-4=0}}} gives solution {{{b=4}}}

{{{b+4=0}}} gives solution {{{b=-4}}} 

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and 

{{{log(3,(5x+5)) - log(3,(x^2 - 1)) = 0}}}

Use this principle on the left:{{{ log(B,(A)) - log(B,(C)) = log(B,(A/C))}}}
 
{{{log(3,(((5x+5))/((x^2-1))))=0}}}

Factor the numerator and denominator:

{{{log(3,((5(x+1))/((x-1)(x+1))))=0}}}

Cancel the {{{(x+1)}}}'s

{{{log(3,((5cross((x+1)))/((x-1)cross((x+1)))))=0}}}

{{{log(3,(5/(x+1)))=0}}}

The log equation {{{log(B,A)=C}}} can be rewritten 
as the exponential equation {{{A=B^C}}}

{{{5/(x-1)=3^0}}}

Any number, except 0, raised to the 0 power is 1. So
we replace {{{3^0}}} by {{{1}}}

{{{5/(x-1)=1}}}

Multiply both side by {{{(x-1)}}}

{{{(x-1)}}}{{{5/(x-1)=(x-1)1}}}

{{{cross((x-1))}}}{{{5/cross((x-1))=(x-1)1}}}

Simplify:

{{{5=x-1}}}

{{{6=x}}}

Edwin</pre>