SHORT ANSWER. Solve the equation and express the solution in exact form using either In or log as indicated. 12) 2^-x=(1)/(8) Express the answer using In. 12) __ 13) ((1)/(4))^x=19 Express the answer using In. 13) __ Use the product, quotient and power rules of logarithms to rewrite the expression as a single logarithm. Assume that all variables represent positive real numbers. 3log_(2)(4x+6)+6log_(2)(3x-7) 14) __

12) To solve the equation $2^{-x}=\frac {1}{8}$, we can rewrite $\frac {1}{8}$ as $2^{-3}$. So, the equation becomes $2^{-x}=2^{-3}$. Since the bases are the same, the exponents must be equal. Therefore, $-x=-3$, which simplifies to $x=3$. To express this in terms of ln, we can take the natural logarithm of both sides: $ln(2^{-x})=ln(\frac {1}{8})$. Using the power rule of logarithms, we get $-x*ln(2)=ln(\frac {1}{8})$. Solving for x, we have $x=-\frac {ln(\frac {1}{8})}{ln(2)}$.<br /><br />13) To solve the equation $(\frac {1}{4})^{x}=19$, we can rewrite $\frac {1}{4}$ as $4^{-1}$. So, the equation becomes $(4^{-1})^{x}=19$, which simplifies to $4^{-x}=19$. Taking the natural logarithm of both sides, we get $ln(4^{-x})=ln(19)$. Using the power rule of logarithms, we have $-x*ln(4)=ln(19)$. Solving for x, we get $x=-\frac {ln(19)}{ln(4)}$.<br /><br />14) To rewrite the expression $3log_{2}(4x+6)+6log_{2}(3x-7)$ as a single logarithm, we can use the product and power rules of logarithms. First, we can factor out the common factor of 3 from the first term: $3log_{2}(4x+6)+6log_{2}(3x-7) = 3log_{2}(4x+6)+2*3log_{2}(3x-7)$. Next, we can use the product rule of logarithms to combine the terms: $3log_{2}(4x+6)+2*3log_{2}(3x-7) = 3log_{2}((4x+6)(3x-7))$. Finally, we can use the power rule of logarithms to simplify the expression: $3log_{2}((4x+6)(3x-7)) = log_{2}((4x+6)(3x-7))^3$.