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Question Number 53205 by mondodotto@gmail.com last updated on 19/Jan/19

Commented by MJS last updated on 22/Jan/19

$$2\mathrm{is}\mathrm{a}\mathrm{solution}$$$$t_1\left(x\right)=\frac{{25}^x}{5^x}={\left(\frac{25}{5}\right)}^x=5^x;t_1\left(2\right)=25$$$$t_2\left(x\right)=5^{x^2-2}=\frac{5^{\left(x^2\right)}}{25};t_2\left(2\right)=25$$$$t_3\left(x\right)=\sqrt{{50}^x}={\left(\sqrt{50}\right)}^x={\left(5\sqrt{2}\right)}^x=5^x{2}^{\frac{x}{2}};t_3\left(x\right)=50$$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$\frac{{25}^x}{5^x}+5^{x^2-2}=\sqrt{{50}^x}$$$$\frac{5^{2x}}{5^x}+5^{x^2-2}={\left(2\times 5^2\right)}^{\frac{x}{2}}$$$$5^x+\frac{5^{x^2}}{5^2}=2^{\frac{x}{2}}\times 5^x$$$$5^x+5^{x^2-2}-2^{\frac{x}{2}}\times 5^x=0$$$$5^x(1+5^{x^2-2}-2^{\frac{x}{2}})=0$$$$if$$$$5^x=0=5^{-\mathrm{\infty }}x=-\mathrm{\infty }$$$$when$$$$1+5^{x^2-2}-2^{\frac{x}{2}}=0$$$$1+5^{x^2-2}=2^{\frac{x}{2}}$$$$nowtakingthehelpofgraph...$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

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