$$\log_{a}{1} = 0$$
$$\log_{a}{xy} = \log_{a}{x}+\log_{a}{y}$$
$$\log_{a}{b}^{c} = c \log_{a}{b}$$
$$\log_{b^n}{a}=\dfrac{1}{n}\log_ba\ $$
$$log_a(x/y) = log_a(x) − log_a(y)$$
$$log_a(a) = 1$$
Basics of logarithm
Very ImportantFormula Video
Question 1
If $$5 - \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 - x} = \log_{10} \frac{1}{\sqrt{1 - x^2}}$$, then 100x equals
Solution
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, $$100\ x\ =100\times\ \frac{99}{100}=99$$
Question 2
For all possible integers n satisfying $$2.25\leq2+2^{n+2}\leq202$$, then the number of integer values of $$3+3^{n+1}$$ is:
Solution
Correct Answer: 7
$$2.25\leq2+2^{n+2}\leq202$$
$$2.25-2\le2+2^{n+2}-2\le202-2$$
$$0.25\le2^{n+2}\le200$$
$$\log_20.25\le n+2\le\log_2200$$
$$-2\le n+2\le7.xx$$
$$-4\le n\le7.xx-2$$
$$-4\le n\le5.xx$$
Possible integers = -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
If we see the second expression that is provided, i.e
$$3+3^{n+1}$$, it can be implied that n should be at least -1 for this expression to be an integer.
So, n = -1, 0, 1, 2, 3, 4, 5.
Hence, there are a total of 7 values.
Question 3
If $$\log_{2}[3+\log_{3} \left\{4+\log_{4}(x-1) \right\}]-2=0$$ then 4x equals
Solution
Correct Answer: 5
We have :
$$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$$
we get $$3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$$
we get $$\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$$
we get $$4+\log_4\left(x-1\right)\ =\ 3$$
$$\log_4\left(x-1\right)\ =\ -1$$
x-1 = 4^-1
x = $$\frac{1}{4}+1=\frac{5}{4}$$
4x = 5
Question 4
For a real number a, if $$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$ then a must lie in the range
Solution
We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .
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