# How To Solve Logarithm Questions Quickly

## Introduction to Logarithm

When  the power of a number must be raised in order to get some other number known as Logarithm. Go through the entire page to know How To Solve Logarithm Questions Quickly. You will come 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.
• Before solving the Logarithm problem the first thing you have to do is take an idea about What is logarithm?
• 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:

1.  logx X= 1
2. loga 1= 0
3. logax =  $\frac{1}{log_x a} \$
4. loga(x p) = p(log ax)
5. loga x=  $\frac{log X}{log a}$
6. loga$\frac{x}{y} \$= loga X- loga Y
7. 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

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

 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.

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

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.