Logarithm problem 3 by Dhaval Furia

Algebra Level 2

2 6 x + 2 3 x + 2 = 21 \large 2^{6x} + 2^{3x + 2} = 21

Find the real root to the equation above.

log ⁡ 2 27 \log_2 27 log ⁡ 2 7 3 \frac {\log_2 7}{3} log ⁡ 2 3 3 \frac {\log_2 3}{3} log ⁡ 2 9 \log_2 9

Dhaval Furia by Dhaval Furia , 26, India

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1 solution

Chew-Seong Cheong
May 22, 2020

2 6 x + 2 3 x + 4 = 21 ( 2 3 x ) 2 + 4 ( 2 3 x ) − 21 = 0 ( 2 3 x − 3 ) ( 2 3 x + 7 ) = 0 ⟹ 2 3 x = 3 Since x is real 3 x log ⁡ 2 = log ⁡ 3 ⟹ x = log ⁡ 3 3 log ⁡ 2 = log ⁡ 2 3 3 \begin{aligned} 2^{6x} + 2^{3x+4} & = 21 \\ (2^{3x})^2 + 4(2^{3x}) - 21 & = 0 \\ \left(2^{3x}-3\right)\left(2^{3x}+7\right) & = 0 \\ \implies 2^{3x} & = 3 & \small \blue{\text{Since }x \text{ is real}} \\ 3x \log 2 & = \log 3 \\ \implies x & = \frac {\log 3}{3\log 2} = \boxed{\frac {\log_2 3}3} \end{aligned}

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