Question

solve the numbers anyone say answer
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Answer 1

b) Solve for x

• $\sf \cfrac{ {2}^{2x - 1} }{ {2}^{x + 2} } = 4$

$\underline{ \green{ \sf {Step \: by \: step \: explanation}}}$

$\implies \sf \cfrac{ {2}^{2x - 1} }{ {2}^{x + 2} } = 4 \\ \\ \implies\sf \cfrac{ {2}^{2x - 1} }{ {2}^{x + 2} } = {2}^{2} \\ \\ \implies \sf {2}^{2x - 1} = {2}^{2} \times {2}^{x + 2} \\ \\ \implies \sf {2}^{2x - 1} = {2}^{2 + x + 2} \\ \\ \implies \sf {2}^{2x - 1} = {2}^{ x + 4} \\ \\ \implies \sf 2x - 1 = x + 4 \\ \\ \implies \sf 2x - x = 4 + 1 \\ \\ \implies \sf x = 5$

$\sf Verification \\→ \sf \cfrac{ {2}^{2x - 1} }{ {2}^{x + 2} } = 4 \\→ \sf \cfrac{ {2}^{2 \times 5- 1} }{ {2}^{5 + 2} } = 4 \\ →\sf \cfrac{ {2}^{10 - 1} }{ {2}^{7} } = 4 \\ →\sf \cfrac{ {2}^{9} }{ {2}^{7} } = 4 \\→ \sf{2}^{9 - 7} = 4 \\ →\sf{2}^{2} = 4 \\→ \sf{4}^{} = 4_{\gray{Verified}}$

Answer 2

Step-by-step explanation:

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