The sum of all the solutions of the equation (8) 2 x -16 ×(8) x +48=0 is :

Mathematics · Quadratic Equations

JEE Main 2024 — 8 April, Shift 1 — Question 2

The sum of all the solutions of the equation $(8)^{2 x}-16 \cdot(8)^{x}+48=0$ is :

$(8)^{2 \mathrm{x}}-16 \cdot(8)^{\mathrm{x}}+48=0$

Put $8^{\mathrm{x}}=\mathrm{t}$

$\mathrm{t}^{2}-16+48=0$

$\Rightarrow \mathrm{t}=4$ or $\mathrm{t}=12$

$\Rightarrow 8^{\mathrm{x}}=4, \quad 8^{\mathrm{x}}=12$

$\Rightarrow \mathrm{x}=\log _{8} \mathrm{X} \quad \mathrm{x}=\log _{8} 12$

sum of solution $=\log _{8} 4+\log _{8} 12$ $=\log _{8} 48=\log _{8}(6.8)$ $=1+\log _{8} 6$

Answer key and solution verified before publishing.

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