[tex]\bf log_3(c+3)-log_3(4c-1)=log_3(5)\\\\
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log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\qquad thus\\\\
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log_3\left( \frac{c+3}{4c-1} \right)=log_3(5)\implies \cfrac{c+3}{4c-1}=5[/tex]
solve for "c"
