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Question Number 167800 by cortano1 last updated on 25/Mar/22

      5^(2x)  = 2.10^x −4^(2x)         (((x−4)^(x^2 +2) ))^(1/((x+2)^(−1) )) =?

$$\:\:\:\:\:\:\mathrm{5}^{\mathrm{2}{x}} \:=\:\mathrm{2}.\mathrm{10}^{{x}} −\mathrm{4}^{\mathrm{2}{x}} \: \\ $$$$\:\:\:\:\:\sqrt[{\left({x}+\mathrm{2}\right)^{−\mathrm{1}} }]{\left({x}−\mathrm{4}\right)^{{x}^{\mathrm{2}} +\mathrm{2}} }=?\: \\ $$

Commented by MJS_new last updated on 25/Mar/22

x=0

$${x}=\mathrm{0} \\ $$

Commented by logan whetzel last updated on 26/Mar/22

please explain this awnser

Commented by MJS_new last updated on 26/Mar/22

5^(2x) +4^(2x) =2×10^x   25^x +16^x =2×10^x   (1) x=0  25^0 +16^0 =2×10^0   1+1=2×1 true  (2) x=−∞  25^(−∞) +16^(−∞) =2×10^(−∞)   0+0=2×0 true    no other solution

$$\mathrm{5}^{\mathrm{2}{x}} +\mathrm{4}^{\mathrm{2}{x}} =\mathrm{2}×\mathrm{10}^{{x}} \\ $$$$\mathrm{25}^{{x}} +\mathrm{16}^{{x}} =\mathrm{2}×\mathrm{10}^{{x}} \\ $$$$\left(\mathrm{1}\right)\:{x}=\mathrm{0} \\ $$$$\mathrm{25}^{\mathrm{0}} +\mathrm{16}^{\mathrm{0}} =\mathrm{2}×\mathrm{10}^{\mathrm{0}} \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{2}×\mathrm{1}\:\mathrm{true} \\ $$$$\left(\mathrm{2}\right)\:{x}=−\infty \\ $$$$\mathrm{25}^{−\infty} +\mathrm{16}^{−\infty} =\mathrm{2}×\mathrm{10}^{−\infty} \\ $$$$\mathrm{0}+\mathrm{0}=\mathrm{2}×\mathrm{0}\:\mathrm{true} \\ $$$$ \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solution} \\ $$

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