Question

PLEASE answer my question fast



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Answer 1

Step-by-step explanation:

(13)

$Given:7^{log\;x} =98 - x^{log\;7}$

$\Longrightarrow 7^{log\;x} +x^{log\;7} = 98$

$\Longrightarrow 2 * 7^{log\;x} = 98$

$\Longrightarrow 7^{log\;x} = 49$

$\Longrightarrow 7^{log\;x} = 7^2$

$\Longrightarrow log\;x = 2$

$\Longrightarrow log\;x = log\; 100$

$\Longrightarrow \boxed{x = 100}$

(14)

$Given:\sqrt{1+log_{0.04}x}+\sqrt{3+log_{0.2}x} = 1$

$\Longrightarrow \sqrt{1+\frac{log_{0.2}(x)}{log_{0.2}(0.04)}} + \sqrt{3+log_{0.2}(x)} = 1$

$\Longrightarrow \sqrt{\frac{2+log_{0.2}(x)}{2}}+\sqrt{3+log_{0.2}(x)} =1$

$Let\;log_{0.02}(x) = u$

$\Longrightarrow \sqrt{\frac{2+u}{2}} + \sqrt{3+u}=1$

$\Longrightarrow \sqrt{\frac{2+u}{2}} = (1-\sqrt{3+u})$

On Squaring both sides, we get

$\Longrightarrow \frac{2+u}{2} = u + 4 - 2\sqrt{3+u}$

$\Longrightarrow \frac{2+u}{2}-u-4 = -2\sqrt{3+u}$

$\Longrightarrow (\frac{2+u}{2}-u-4)^2 = (-2\sqrt{3+u})^2$

$\Longrightarrow \frac{u^2}{4}+3u+9=12+4u$

$\Longrightarrow u^2 - 4u - 12 = 0$

$\Longrightarrow u^2 -6u + 2u - 12 = 0$

$\Longrightarrow u(u-6)+2(u-6)=0$

$\Longrightarrow (u+2)(u-6)=0$

$\Longrightarrow u = -2,6[Not \;satisfies]$

$\Longrightarrow u = -2$

$\Longrightarrow log_{0.2}(x) = -2$

$\Longrightarrow log_{0.2}(x) = log_{0.2}(\frac{1}{0.04})$

$\Longrightarrow x = \frac{1}{0.04}$

$\Longrightarrow \boxed{x = 25}$

Hope it helps!