Question 644599

Solve. log(base4)(x+4)-log(base16)x=1


{{{log_[4] (x + 4) - log_[16] (x) = 1}}}
{{{log_[4] (x + 4) - log_[4] x/log_[4] 16 = 1}}} ----- Applying change of base to base 4
{{{log_[4] (x + 4) - (log_[4]x)/2 = 1}}}
{{{2*log_[4] (x + 4) - log_[4] x = 2}}} ----- Multiplying by LCD, 2
{{{log_[4] (x + 4)^2 - log_[4] x = 2}}}
{{{log_[4] ((x + 4)^2/x) = 2}}}
{{{(x + 4)^2/x = 4^2}}}
{{{(x + 4)^2 = 16x}}} ----- Cross-multiplying
{{{x^2 + 8x + 16 = 16x}}}
{{{x^2 + 8x - 16x + 16 = 0}}}
{{{x^2 - 8x + 16 = 0}}}


(x – 4)(x – 4) = 0

{{{highlight_green(x = 4)}}}


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Check
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{{{log_[4] (4 + 4) - log_[16] (4) = 1}}}
{{{log_[4] 8 - log_[16] 4 = 1}}}
1.5 - .5 = 1
1 = 1 (TRUE)


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