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Question

Mathematics Question on Sequence and series

If loge5log_e\,5, loge(5x1)log_e(5^x-1) and logelog_e (5x115)\left(5^{x}-\frac{11}{5}\right) are in A.P.A.P., then the values of xx are

A

log54\log_5 \,4 and log53\log_5\, 3

B

log34\log_3\, 4 and log43\log_4\, 3

C

log34\log_3\, 4 and log35\log_3\, 5

D

log56\log_5\,6 and log57\log_5\, 7

Answer

log54\log_5 \,4 and log53\log_5\, 3

Explanation

Solution

Since, loge5,loge(5x1)\log _{e} 5, \log _{e}\left(5^{x}-1\right) and loge(5x115)\log _{e}\left(5^{x}-\frac{11}{5}\right)
are in AP.
2loge(5x1)=loge5+loge(5x115)\therefore 2 \log _{e}\left(5^{x}-1\right)=\log _{e} 5+\log _{e}\left(5^{x}-\frac{11}{5}\right)
(5x1)2=5(5x115)\Rightarrow \left(5^{x}-1\right)^{2}=5\left(5^{x}-\frac{11}{5}\right)
52x+12×5x=5×5x11\Rightarrow 5^{2 x}+1-2 \times 5^{x}=5 \times 5^{x}-11
52x7×5x+12=0\Rightarrow 5^{2 x}-7 \times 5^{x}+12=0
52x4×5x3×5x+12=0\Rightarrow 5^{2 x}-4 \times 5^{x}-3 \times 5^{x}+12=0
(5x4)(5x3)=0\Rightarrow \left(5^{x}-4\right)\left(5^{x}-3\right)=0
5x=4,5x=3\Rightarrow 5^{x}=4,5^{x}=3
x=log54,x=log53\Rightarrow x=\log _{5} 4, x=\log _{5} 3