Alg9.4.13

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Find the value of x if $\log_4(x^2+x)-\log_4(x+1)=2\,$

By applying $\log_a(m)-\log_a(n)=\log_a(\frac{m}{n})\,$

we get

$\log_4(\frac{x^2+x}{x+1})=2\,$

$\log_4(\frac{x(x+1}{x+1})=2\,$

By simplifying the above equation,

$\log_4(x)=2\,$

This can be expressed as

$x=4^2\,$

Hence the value of x=4

Main Page:Algebra:Logarithmic

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