((8^x −2^x )/(6^x −3^x ))=2 , find x ? Mastermind


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Question Number 170032 by Mastermind last updated on 14/May/22

$\frac{8^{x} - 2^{x}}{6^{x} - 3^{x}} = 2, f i n d x ?$ $M a s t e r m i n d$
Commented by Mastermind last updated on 14/May/22

$y o u r s o l u t i o n p l e a s e$
Commented by Rasheed.Sindhi last updated on 14/May/22

$x = 1$
	Answered by Rasheed.Sindhi last updated on 14/May/22

	$\frac{8^{x} - 2^{x}}{6^{x} - 3^{x}} = 2, f i n d x ?$ $\frac{2^{x} (4^{x} - 1)}{3^{x} (2^{x} - 1)} = 2$ $\frac{2^{x} (2^{2 x} - 1)}{3^{x} (2^{x} - 1)} = 2$ $\frac{2^{x} (2^{x} - 1) (2^{x} + 1)}{3^{x} (2^{x} - 1)} = 2; 2^{x} \neq 1 \Rightarrow x \neq 0^{★}$ $\frac{2^{x} (2^{x} + 1)}{3^{x}} = 2$ $A s s u m i n g x a w h o l e n u m b e r$ $2^{x} \cdot \frac{2^{x} + 1}{3^{x}} = 2 = 1 \times 2$ $(2^{x} = 1 \land \frac{2^{x} + 1}{3^{x}} = 2) \lor (2^{x} = 2 \land \frac{2^{x} + 1}{3^{x}} = 1)$ $(2^{x} = 2^{0} \land \frac{2^{0} + 1}{3^{0}} = 2) \lor (2^{x} = 2^{1} \land \frac{2^{1} + 1}{3^{1}} = 1)$ $x \neq 0^{★}$ $H e n c e x = 1 (O n l y o n e s o l u t i o n i n w h o l e n u m b e r s)$
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