This is a basic logarithm problem

Number Theory Level 2

log ⁡ 6 2 x + 3 − log ⁡ 6 ( 3 x − 2 ) = x \log_6 2^{x+3} - \log_6 (3^x - 2) = x

Find x x .

log ⁡ 3 5 \log_3 5 log ⁡ 3 4 \log_3 4 log ⁡ 3 16 \log_3 16 log ⁡ 3 7 \log_3 7

Prakhar Verma by Prakhar Verma , 18, India

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1 solution

Chew-Seong Cheong
Apr 19, 2020

log ⁡ 6 2 x + 3 − log ⁡ 6 ( 3 x − 2 ) = x log ⁡ 6 ( 2 x + 3 3 x − 2 ) = log ⁡ 6 6 x 8 ⋅ 2 x 3 x − 2 = 2 x ⋅ 3 x ⟹ ( 3 x ) 2 − 2 ( 3 x ) − 8 = 0 ( 3 x + 2 ) ( 3 x − 4 ) = 0 Since 3 x > 0 3 x = 4 ⟹ x = log ⁡ 3 4 \begin{aligned} \log_6 2^{x+3} - \log_6 (3^x-2) & = x \\ \log_6 \left(\frac {2^{x+3}}{3^x-2} \right) & = \log_6 6^x \\ \frac {8\cdot 2^x}{3^x-2} & = 2^x \cdot 3^x \\ \implies (3^x)^2 - 2(3^x) - 8 & = 0 \\ (3^x+2)(3^x - 4) & = 0 & \small \blue{\text{Since }3^x > 0} \\ 3^x & = 4 \\ \implies x & = \boxed{\log_3 4} \end{aligned}

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