Question 1094663
<pre>
{{{log((x+14))+log((x+2))}}}{{{""=""}}}{{{6}}}

Use the principle: {{{log(B,(A))+log(B,(C))=log(B,(A*C))}}} to write
the left side as a single logarithm:

{{{log(((x+14)^""(x+2)))}}}{{{""=""}}}{{{6}}}

Use the principle: {{{log((A))=log(10,(A))}}} to put in the understood
logarithm base:

{{{log(10,((x+14)^""(x+2)))}}}{{{""=""}}}{{{6}}}

Use the principle that the logarithm equation {{{log(B,(A))=C}}} is equivalent
to the exponential equation {{{A=B^C}}} to rewrite the above logatithm
equation as an exponential equation:

{{{(x+14)(x+2)}}}{{{""=""}}}{{{10^6}}}

{{{x^2+16x+28)}}}{{{""=""}}}{{{1000000}}}

{{{x^2+16x-999972)}}}{{{""=""}}}{{{0}}}

Substitute a=1, b=16, and c=-999972 in the quadratic formula:

{{{x}}}{{{""=""}}}{{{(-b +- sqrt( b^2-4ac ))/(2a) }}}

{{{x}}}{{{""=""}}}{{{(-(16) +- sqrt((16)^2-4(1)(-999972)))/(2(1)) }}}

{{{x}}}{{{""=""}}}{{{(-16 +- sqrt(256+3999888))/2 }}}

{{{x}}}{{{""=""}}}{{{(-16 +- sqrt(4000144))/2 }}}

{{{x}}}{{{""=""}}}{{{(-16 +- sqrt(16*250009))/2 }}}

{{{x}}}{{{""=""}}}{{{(-16 +- 4sqrt(250009))/2 }}}

{{{x}}}{{{""=""}}}{{{(-16)/2 +- (4sqrt(250009))/2 }}}

{{{x}}}{{{""=""}}}{{{-8 +- 2sqrt(250009) }}}

We discard the negative answer since logarithms of negative
numbers are not real.  So the only solution is

{{{x}}}{{{""=""}}}{{{-8 + 2sqrt(250009) }}}

That's approximately 992.0179998

Edwin</pre>