Question 740355
{{{log(4,(x+4)) + log(4,(x+4)) = 1}}}....in base {{{10}}} we have


{{{2log((x+4))/log(4) = 1}}}


{{{2log((x+4)) = 1*log(4)}}}


{{{log((x+4)^2) = log(4)}}}...if so, then


{{{(x+4)^2 = 4}}}

{{{x^2+8x+16 = 4}}}

{{{x^2+8x+16 -4=0}}}

{{{x^2+8x+12=0}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-8 +- sqrt( 8^2-4*1*12 ))/(2*1) }}}

{{{x = (-8 +- sqrt( 64-48 ))/2 }}}

{{{x = (-8 +- sqrt( 16))/2 }}}

{{{x = (-8 +- 4)/2 }}}

solutions;

{{{x = (-8 +-4)/2 }}}

{{{x = -4/2 }}}

{{{highlight(x = -2) }}}.......


{{{x = (-8 -4)/2 }}}

{{{x = -12/2 }}}

{{{x = -6 }}}


we need only this solution because we have {{{logarithm}}} and we can find only log of positive number; since we looking for log of{{{(x+4)}}} it will be positive number if we plug in {{{x=-2}}} ({{{(x+4)=-2+4=2}}})


if we plug in {{{x=-6}}} we will get negative number ({{{(x+4)=-6+4=-2}}})

so, your solution is: {{{highlight(x = -2) }}}


{{{ graph( 600, 600, -10, 10, -10, 10, log(4,(x+4)) + log(4,(x+4)),2log((x+4))/log(4), 1) }}}