# solve the equation log(base16)(3x-1)= log(base4)(3x)+log(base4)0.5?

solve the equation $\mathrm{log}\left(base16\right)\left(3x-1\right)=\mathrm{log}\left(base4\right)\left(3x\right)+\mathrm{log}\left(base4\right)0.5$?

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unessodopunsep

l$og\left[4\right]\left(16\right)\cdot \mathrm{log}\left[16\right]4=1$ where [] denotes base, is an example of a fundamental rule in logarithms: $\mathrm{log}\left[a\right]\left(b\right)\cdot \mathrm{log}\left[b\right]\left(a\right)=1$, so, since $\mathrm{log}\left[4\right]\left(16\right)=2\mathrm{log}\left[4\right]4=2,\mathrm{log}\left[16\right]4=\frac{1}{2}$, or, alternatively, $log\left[16\right]4=log\left[16\right]\frac{{16}^{1}}{2}=\frac{1}{2}log\left[16\right]16=\frac{1}{2}.$
Also, $\mathrm{log}\left[a\right]\left(x\right)=\mathrm{log}\left[b\right]\frac{x}{\mathrm{log}\left[b\right]}$
(a). $\mathrm{log}\left[16\right]\left(3x-1\right)=\mathrm{log}\left[4\right]\left(3x\right)+\mathrm{log}\left[4\right]\left(0.5\right)⇒2\mathrm{log}\left[4\right]\left(3x-1\right)=\mathrm{log}\left[4\right]\left(3x\cdot 0.5\right)$. We can equate the logs: ${\left(3x-1\right)}^{2}=3\frac{x}{2}\to 9{x}^{2}-6x+1=3\frac{x}{2}⇒18{x}^{2}-12x-3x+2=0$
$\to 18{x}^{2}-15x+2=0\to \left(3x-2\right)\left(6x-1\right)=0$
From this $x=\frac{2}{3}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{1}{6}.$ Substitute these values in the original equation: $0=\frac{1}{2}-\frac{1}{2}$. The value $\frac{1}{6}$ cannot be used because it would require the log of a negative number so the only solution is $Lx=\frac{2}{3}$