# 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. a loglogax =X

4.logax =  $\frac{1}{log_x a} \$

5. loga(x p) = p(log ax)

6. loga x= $\frac{ log_a X }{ log_b a } \$ = $\frac{logX}{log_a} \$

7. loga$\frac{x}{y} \$= loga X- logb Y

8. 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

#### Explanation:

log 27= 1.431

$\implies$ log(3)3= 1.431
$\implies$ 3log 3= 1.431
$\implies$ log3= 0.477

therefore, log9= log(32= 2log3= (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

#### Explanation

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

## Solve Quickly Logarithm Questions

### Question 3

Solve the equation log x= 1- log(x-3)

#### Explanation:

By combining both the equation we get

logx + log (x-3)=1

log(x(x-3))= 1

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

#### Explanation:

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)=1
1+ 2log2 x= 3
2log2 x = 2
log2x = 1
x= 2

### Question 5

Iflog 10 5+ log(5x+1) = log 10 (x+5) +1, Find the value of X?

Option

A) 3
B) 1
C) 10
D) 5 