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PREPINSTA PRIME
How To Solve Logarithm Questions Quickly
Introduction to Logarithm
Go through the entire page to know How To Solve Logarithm Questions Quickly. You will get to know easy tricks and formulas of Logarithms problems.
How To Solve Logarithm Question Quickly:-
- One of the most important and marks scoring section in Aptitude is Logarithm. Logarithm might be confusing but once you start solving it; it will become one of the easiest topics you have ever solved.
- This page will help you out in solving the problems regarding Logarithm.
- Before solving the Logarithm problem the first thing you have to do is take an idea about What is logarithm?
- A logarithm is a power to which a number must be raised to get another number.
- There are some basic formulas which are required to Solve Logarithm Questions quickly.
Logarithm Formulas:
- logx X= 1
- loga 1= 0
- logax = \frac{1}{log_x a} \
- loga(x p) = p(log ax)
- loga x= \frac{log X}{log a}
- loga \frac{x}{y} \ = loga X- loga Y
- loga (xy)= logaX+ logaY
Question 1 If log 27= 1.431, then the value of log 9 is?
Option:
A) 0.945
B) 0.934
C) 0.958
D) 0.954
Solution: log 27= 1.431
\implies log(3)3= 1.431
\implies 3log 3= 1.431
\implies log3= 0.477
therefore, log9= log 32= 2 log3= (2×0.477)= 0.954
Correct Answer: D
Question 2 If log \frac{a}{b} \ +log \frac{b}{a} \ = log(a+b), then
Option:
A) a-b=1
B) a=b
C) a+b=1
D)a2-b2 = 1
Solution: log \frac{a}{b} \ + log \frac{b}{a} \ = log(a+b)
\implies log (a+b)= log ( \frac{a}{b} \ x \frac{b}{a}) \ = log 1
so, a+b=1
Correct Answer: C
Exponential Form | Logarithm Formula |
a x = y | x = log a y |
(a m ) n | log m n = n × log m |
a m × a n = a m+n | log (m × n) = log m + log n |
Best way Quickly Solve Logarithm Questions
Question: Solve the equation log x= 1- log(x-3)
Option:
A) 2
B) \frac{1}{2} \
C) 5
D) 4
Solution: By combining both the equation we get
logx + log (x-3)=1
log(x(x-3))= log 101
Now convert it into exponential form,
x (x-3)= 101
x2 – 3x-10= 0
(x-5) (x+2)=0
x= -2, x=5
By solving this equation we get two values for x.
x= -2, x=5
Put the different value of x in different equation and solve them,
x= -2
log(-2) = 1- log (-2-3)
x= 5
log5 = 1-log(5-3)
log5 = 1-log2
Negative value is not considered in logarithm. So, we have a single value of x i.e, x=5.
Correct Answer : C.
Question 4 log9 (3log2 (1+log3 (1+2log2 x)))= \frac{1}{2} \ . Find x.
Option:
A) 2
B) \frac{1}{2} \
C) 1
D) 4
Solution : log9 (3log2 (1+log3 (1+2log2 x)))= \frac{1}{2} \ .
3log2 (1+log3 (1+2log2 x))= 9^{\frac{1}{2}} = 3
log2(1+log3(1+2log2 x) = 1
1+log3 (1+2log2 x)=2^1
log3 (1+2log2 x)=2^1 -1
log3 (1+2log2 x) = 1
(1+2log2 x) = 3^1
1+ 2log2 x= 3
2log2 x = 2
log2x = 1
x= 2
Correct Answer: A
Question 5 If log 10 5+ log(5x+1) = log 10 (x+5) +1, Find the value of X?
Option
A) 3
B) 1
C) 10
D) 5
Solution : log 10 5+ log(5x+1) = log 10 (x+5) +1.
log 10 5+ log(5x+1) = log 10 (x+5) +log 10 10
log10 [5 (5x+1) ] = log10( 10 (x+5)]
5 (5x+1) = 10 (x+5)
5x+1 = 2x+ 10
3x= 9
x=3.
Correct Answer: A
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