# How To Solve Logarithm Questions Quickly

### 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 problems.

### How To Solve Quickly Question on Logarithm

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- logb Y
7. loga (xy)= logaX+ logbY

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} \$ + $\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.