solve algebraically

**joeyisnice123** · November 20th 2008, 08:18 PM

2log(base4)X - log(base4)(x+3)=1

**Chris L T521** · November 20th 2008, 08:49 PM

Originally Posted by joeyisnice123

2log(base4)X - log(base4)(x+3)=1

Note that $2\log_4x-\log_4\left(x+3\right)=1\implies \log_4\left(x^2\right)-\log_4\left(x+3\right)=1$

Now recall that $\log_ab-\log_ac=\log_a\left(\frac bc\right)$

Thus, $\log_4\left(x^2\right)-\log_4\left(x+3\right)=1\implies\log_4\left(\frac{ x^2}{x+3}\right)=1$

Now, note that $\log_44=1$

Thus, we now have the equation $\log_4\left(\frac{x^2}{x+3}\right)=\log_44$

Since we want the two logarithms to be equal, we must see where $\frac{x^2}{x+3}=4$

Can you continue on from here?

--Chris

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