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Question Number 166431 by BagusSetyoWibowo last updated on 20/Feb/22

2(2^x /(16,0)) = 5,7^2  − 0,49 + 0  How much the x is?

$$\mathrm{2}\frac{\mathrm{2}^{{x}} }{\mathrm{16},\mathrm{0}}\:=\:\mathrm{5},\mathrm{7}^{\mathrm{2}} \:−\:\mathrm{0},\mathrm{49}\:+\:\mathrm{0} \\ $$$${How}\:{much}\:{the}\:{x}\:{is}? \\ $$

Answered by alephzero last updated on 20/Feb/22

2(2^x /(16)) = 32  ⇒ ((32+2^x )/(16)) = 32  ⇒ 2^x  = 480  ⇒ x = log_2  480  ⇒ x = log_2  (32 ∙ 15)  ⇒ x = log_2  32 + log_2  15  ⇒ x = 5 + log_2  15

$$\mathrm{2}\frac{\mathrm{2}^{{x}} }{\mathrm{16}}\:=\:\mathrm{32} \\ $$$$\Rightarrow\:\frac{\mathrm{32}+\mathrm{2}^{{x}} }{\mathrm{16}}\:=\:\mathrm{32} \\ $$$$\Rightarrow\:\mathrm{2}^{{x}} \:=\:\mathrm{480} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{log}_{\mathrm{2}} \:\mathrm{480} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{32}\:\centerdot\:\mathrm{15}\right) \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{log}_{\mathrm{2}} \:\mathrm{32}\:+\:\mathrm{log}_{\mathrm{2}} \:\mathrm{15} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{5}\:+\:\mathrm{log}_{\mathrm{2}} \:\mathrm{15} \\ $$

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