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Question Number 223615 by fantastic last updated on 31/Jul/25
40^(x−1) =2^(2x+1)
$$\mathrm{40}^{{x}−\mathrm{1}} =\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$
Answered by mr W last updated on 31/Jul/25
((40^x )/(40))=2×2^(2x) =2×4^x 10^x =80 ⇒x=log 80=1+log 8≈1.903
$$\frac{\mathrm{40}^{{x}} }{\mathrm{40}}=\mathrm{2}×\mathrm{2}^{\mathrm{2}{x}} =\mathrm{2}×\mathrm{4}^{{x}} \\ $$$$\mathrm{10}^{{x}} =\mathrm{80} \\ $$$$\Rightarrow{x}=\mathrm{log}\:\mathrm{80}=\mathrm{1}+\mathrm{log}\:\mathrm{8}\approx\mathrm{1}.\mathrm{903} \\ $$
Answered by Raphael254 last updated on 02/Aug/25
40^(x−1) = 2^(2x+1) (2^n )^(x−1) = 2^(2x+1) 2^(nx − n) = 2^(2x+1) nx − n = 2x + 1 nx − 2x = 1 + n x(n − 2) = 1 + n x = ((1 + n)/(n − 2)) = ((n + 1)/(n − 2)) 2^n = 40 ⇒ log_2 40 = n n = log_2 (8×5) = log_2 8 + log_2 5 n = log_2 2^3 + log_2 (4×(5/4)) n = 3 + log_2 4 + log_2 (5/4) n = 3 + 2 + ≈((19)/(59)) n = 5 + ≈((19)/(59)) n ≈ ((295 + 19)/(59)) ≈ ((314)/(59)) x = ((n + 1)/(n − 2)) ⇒ x ≈ ((((314)/(59)) + 1)/(((314)/(59)) − 2)) x ≈ (((314 + 59)/(59))/((314 − 118)/(59))) ≈ (((373)/(59))/((196)/(59))) ≈ ((373)/(196)) ≈ 1.903
$$ \\ $$$$\mathrm{40}^{{x}−\mathrm{1}} \:=\:\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\left(\mathrm{2}^{{n}} \right)^{{x}−\mathrm{1}} \:=\:\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\mathrm{2}^{{nx}\:−\:{n}} \:=\:\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$$${nx}\:−\:{n}\:=\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$$${nx}\:−\:\mathrm{2}{x}\:=\:\mathrm{1}\:+\:{n} \\ $$$${x}\left({n}\:−\:\mathrm{2}\right)\:=\:\mathrm{1}\:+\:{n} \\ $$$${x}\:=\:\frac{\mathrm{1}\:+\:{n}}{{n}\:−\:\mathrm{2}}\:=\:\frac{{n}\:+\:\mathrm{1}}{{n}\:−\:\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{2}^{{n}} \:=\:\mathrm{40}\:\Rightarrow\:{log}_{\mathrm{2}} \:\mathrm{40}\:=\:{n} \\ $$$$ \\ $$$${n}\:=\:{log}_{\mathrm{2}} \:\left(\mathrm{8}×\mathrm{5}\right)\:=\:{log}_{\mathrm{2}} \:\mathrm{8}\:+\:{log}_{\mathrm{2}} \:\mathrm{5} \\ $$$${n}\:=\:{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{3}} \:+\:{log}_{\mathrm{2}} \:\left(\mathrm{4}×\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$${n}\:=\:\mathrm{3}\:+\:{log}_{\mathrm{2}} \:\mathrm{4}\:+\:{log}_{\mathrm{2}} \:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${n}\:=\:\mathrm{3}\:+\:\mathrm{2}\:+\:\approx\frac{\mathrm{19}}{\mathrm{59}} \\ $$$${n}\:=\:\mathrm{5}\:+\:\approx\frac{\mathrm{19}}{\mathrm{59}} \\ $$$${n}\:\approx\:\frac{\mathrm{295}\:+\:\mathrm{19}}{\mathrm{59}}\:\approx\:\frac{\mathrm{314}}{\mathrm{59}} \\ $$$$ \\ $$$${x}\:=\:\frac{{n}\:+\:\mathrm{1}}{{n}\:−\:\mathrm{2}}\:\Rightarrow\:{x}\:\approx\:\frac{\frac{\mathrm{314}}{\mathrm{59}}\:+\:\mathrm{1}}{\frac{\mathrm{314}}{\mathrm{59}}\:−\:\mathrm{2}} \\ $$$${x}\:\approx\:\frac{\frac{\mathrm{314}\:+\:\mathrm{59}}{\mathrm{59}}}{\frac{\mathrm{314}\:−\:\mathrm{118}}{\mathrm{59}}}\:\approx\:\frac{\frac{\mathrm{373}}{\mathrm{59}}}{\frac{\mathrm{196}}{\mathrm{59}}}\:\approx\:\frac{\mathrm{373}}{\mathrm{196}}\:\approx\:\mathrm{1}.\mathrm{903} \\ $$
Answered by Frix last updated on 01/Aug/25
(x−1)ln 40 =(2x+1)ln 2 xln 40 −2xln 2=ln 40 +ln 2 x(ln 40 −2ln 2)=ln 40 +ln 2 x(ln 5 +ln 2)=ln 5 +4ln 2 x=((ln 5 +4ln 2)/(ln 5 +ln 2))
$$\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{40}\:=\left(\mathrm{2}{x}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{2} \\ $$$${x}\mathrm{ln}\:\mathrm{40}\:−\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}=\mathrm{ln}\:\mathrm{40}\:+\mathrm{ln}\:\mathrm{2} \\ $$$${x}\left(\mathrm{ln}\:\mathrm{40}\:−\mathrm{2ln}\:\mathrm{2}\right)=\mathrm{ln}\:\mathrm{40}\:+\mathrm{ln}\:\mathrm{2} \\ $$$${x}\left(\mathrm{ln}\:\mathrm{5}\:+\mathrm{ln}\:\mathrm{2}\right)=\mathrm{ln}\:\mathrm{5}\:+\mathrm{4ln}\:\mathrm{2} \\ $$$${x}=\frac{\mathrm{ln}\:\mathrm{5}\:+\mathrm{4ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{5}\:+\mathrm{ln}\:\mathrm{2}} \\ $$
Commented by fantastic last updated on 01/Aug/25
thank you all
$${thank}\:{you}\:{all} \\ $$

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