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(B) Given the equations are of the form $a^x = \frac{9}{5}$,where $a$ is the base.Taking the logarithm on both sides,$x \ln(a) = \ln(1.8)$,which impli

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$5\left(1+\frac{r}{17}\right)^x=9$

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(B) Given the equations are of the form $a^x = \frac{9}{5}$,where $a$ is the base.Taking the logarithm on both sides,$x \ln(a) = \ln(1.8)$,which implies $x = \frac{\ln(1.8)}{\ln(a)}$.Since $\ln(1.8) > 0$,$x$ is largest when $\ln(a)$ is smallest and positive,which occurs when the base $a$ is closest to $1$ (but greater than $1$).Let the bases be $a_1 = 1 + \frac{r}{\pi}$,$a_2 = 1 + \frac{r}{17}$,$a_3 = 1 + 2r$,and $a_4 = 1 + \frac{1}{r}$.Given $0 $a_2 = 1 + \frac{r}{17}$ is the smallest value because $\frac{r}{17}$ is the smallest increment among the options for $r \in (0, 4)$.Since $a_2$ is the smallest base greater than $1$,the exponent $x$ required to reach the value $\frac{9}{5}$ must be the largest.Therefore,option $B$ is correct.

Below are four equations in $x$. Assume that $0 < r < 4$. Which of the following equations has the largest solution for $x$?

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