Surds and their properties

$$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$

We can re-write the equation as: $$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$$

$$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$$

$$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$$

$$5=-5\log_{10}\sqrt{1-x}$$

$$\sqrt{1-x}=\frac{1}{10}$$

Squaring both sides: $$\left(\sqrt{1-x}\right)^2=\frac{1}{100}$$

$$\therefore\ $$ $$x=1-\frac{1}{100}=\frac{99}{100}$$

Hence, $$100\ x\ =100\times\ \frac{99}{100}=99$$

$$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{125}{343}$$

$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$$

3x-y = -6

$$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$

Therefor, y=6x as the bases are different so the power should be zero for the results to be equal.

3x-y=-6

or, 3x - 6x = -6

or x= 2

y= 6x = 12

x+y = 14

Given, $$\log_{x}(x^2 + 12) = 4$$

=> $$x^2+12=x^4$$

=> $$x^4-x^2-12=0$$

=> $$x^4-4x^2+3x^2-12=0$$

=> $$x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$$

=> $$\left(x^2-4\right)\left(x^2+3\right)=0$$ => since, x is a positive real number (given) => x = 2.

Now, Given $$3 \log_{y} x = 1$$

=> $$\log_yx=\frac{1}{3}$$

=> $$x=y^{\frac{1}{3}}$$

=> $$y=x^3$$ => y = 8.

=> x + y = 2 + 8 = 10.