$$a^m\times\ a^n=a^{m+n}$$
$$\frac{a^m\ \ }{a^n}\ =a^{m-n}$$
$$\left(a^m\right)^{^n}=a^{m\times\ n}$$
$$\left(a\times\ b\right)^m\ =a^m\times\ b^m$$
$$a^{-m}=\ \frac{1}{a^m}$$
$$a^{\frac{m}{n}}=\sqrt[\ n]{a^m}$$
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Check Now$$a^m\times\ a^n=a^{m+n}$$
$$\frac{a^m\ \ }{a^n}\ =a^{m-n}$$
$$\left(a^m\right)^{^n}=a^{m\times\ n}$$
$$\left(a\times\ b\right)^m\ =a^m\times\ b^m$$
$$a^{-m}=\ \frac{1}{a^m}$$
$$a^{\frac{m}{n}}=\sqrt[\ n]{a^m}$$
If $$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$, then 100x equals
Correct Answer: 99
$$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$
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, $$100\ x\ =100\times\ \frac{99}{100}=99$$
If $$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$ and $$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$, for all non-zero real values of a and b, then the value of $$x+y$$ is
Correct Answer: 14
$$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$
$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{125}{343}$$
$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$$
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
If $$x$$ and $$y$$ are positive real numbers such that $$\log_{x}(x^2 + 12) = 4$$ and $$3 \log_{y} x = 1$$, then $$x + y $$ equals
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.
If $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$, then $$\sqrt{10x+9}$$ is equal to
Given, $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$
=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=6+3\sqrt{\ 2}$$
=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=\sqrt{\ 36}+\sqrt{\ 18}$$
Comparing the L.H.S. and R.H.S.
=> $$5x+9=36\ $$ => $$5x=27$$ => $$x=\dfrac{27}{5}$$ (can be verified using the second term as well).
=> $$\sqrt{10x+9}$$ = $$\sqrt{\left(10\times\dfrac{27}{5}\right)+9}$$ = $$\sqrt{\ 63}=3\sqrt{\ 7}$$
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