Question

If $lo{g}_{e}5$, $lo{g}_e(5^x-1)$ and $lo{g}_e$ $\left(5^x-\frac{11}{5}\right)$ are in $A.P.$, then the values of $x$ are

• A. ${\log }_{5}4$ and ${\log }_{5}3$
• B. ${\log }_{3}4$ and ${\log }_{4}3$
• C. ${\log }_{3}4$ and ${\log }_{3}5$
• D. ${\log }_{5}6$ and ${\log }_{5}7$
• E. 12 ,6

Answer 1

Answer: (A)

Since, ${\log }_{e}5,{\log }_e\left(5^x-1\right)$ and ${\log }_e\left(5^x-\frac{11}{5}\right)$
are in AP.
$\therefore 2{\log }_e\left(5^x-1\right)={\log }_{e}5+{\log }_e\left(5^x-\frac{11}{5}\right)$
$\Rightarrow {\left(5^x-1\right)}^2=5\left(5^x-\frac{11}{5}\right)$
$\Rightarrow 5^{2x}+1-2\times 5^x=5\times 5^x-11$
$\Rightarrow 5^{2x}-7\times 5^x+12=0$
$\Rightarrow 5^{2x}-4\times 5^x-3\times 5^x+12=0$
$\Rightarrow \left(5^x-4\right)\left(5^x-3\right)=0$
$\Rightarrow 5^x=4,5^x=3$
$\Rightarrow x={\log }_{5}4,x={\log }_{5}3$