## Problems on logarithms for aptitude

Let’ move on to problems on logarithms for aptitude to develop a better understanding of these concepts-

**Problem-1** Find the value of

**Options-** A) log_{n}32!

B) log_{n}22!

C) log_{n}43!

D) log_{n}43

**Answer-** C) log_{n}43!

**Solution-** Since the =

Using the formulae that

Thus, option (C)! is the correct answer.

**Problem-2** Find the value of x, if

**Options-** A) x=2

B) x=√5

C) x=√3

D) x=√6

**Answer-** C) x=√5

**Solution-** Given that,

=>

Now, we can use the exponent rule,

=>RHS= -Eqn-1

Also, LHS= -Eqn-2

On comparing Eqn.1 and Eqn.2

=>2+√5=2+x

=> x=√5

Thus, the option (C) x=√5 is the correct answer.

**Problem-3** Find the value of x, if

**Options-** A) x= 1⁄12

B) x=6

C) x=2⁄17

D) x=3⁄17

**Answer- **C) x=2⁄17

**Solution- **Given that,

Since we know that,

=>

And also,

=>

On comparing both sides of the equation, we get-

=>

Thus, the option(C) x=2⁄17 is the correct answer.

**Problem-4** Find the relation between a, b, c, if

**Options-** A) a^{2}+b^{2}=c^{2}

B) a^{2}-b^{2}=c^{2}

C) a^{2}+c^{2}=b^{2}

D) a^{2}+b^{2}+c^{2}=0

**Answer-** (A) a^{2}+b^{2}=c^{2}

**Solution- **Given that,

Since we know that

=>

Now, using the multiplication rule,

=>

Now, we can separate the variables using the property-.

Again by the exponent rule, we know that

On comparing the above logarithms we can conclude that

=>c^{2}+a^{2}=b^{2}

Thus, option (B)c^{2}+a^{2}=b^{2 }is the correct answer.

**Problem-5** Find the value of x^{7}.y^{4}.z^{-5}, if

**Options-** A)- 2

B)- 1

C)- 3

D)- 4

**Answer-** (B)- 1

**Solution- **Since it is given that,

Let, there be a constant k, such that

Now we need to find the value of x^{7}.y^{4}.z^{--5}

So, let’s suppose x^{7}.y^{4}.z^{-5}= Q

Taking log of both sides,

Since, we know that

Thus, the option(B) x^{7}.y^{4}.z^{-5}=1 is correct option.

**Problem-6** is equal to

**Options- **A)- 2

B)- 3

C)- 4

D)- 5

**Answer- **(A)- 2

**Solution- **We need to evaluate

Since the base of every term is different so we can use the base change formulae, here i.e.,

=> =

Since, 25=5^{2}, 9=3^{2}, 16=4^{2}, we can use the exponent rule, i.e.,

Thus, option(A)- 2 is the correct option.

**Problem-7** If Find the value of

**Options- **A)- 2y+x-(¼)z

B)- 2y-x+(¼)z

C)- y+3x+(¼)z

D)- 2y+x+(¼)z

**Answer-** (D)- 2y+x+(¼)z

**Solution- **It is given that and we need to find the value of , we will start with simplifying the term using the exponent formulae and the multiplication formulae.

^{)}

On substituting the values given in the questions

Thus, the option(D)2y+x+(¼)z is the correct answer.

**Problem-8** Find the value of

**Options- **A)- 2

B)- 1

C)- 3

D)- 4

**Answer- ** (B)- 1

Solution- To find the value of ,

We will start with replacing 1 in the denominators using the following formulae i.e.,

Now, we can use the multiplication formulae,

Now, we can use the base reverse formulae to simplify the expression i.e.,

Now, we can use the multiplication formulae to further simplify the expression i.e.,

=

Thus, option (B)- 1 is the correct answer.

## Frequently Asked Questions

- What is the multiplication rule in logarithms?

The multiplication rule in the logarithm allows us to operate on the logarithm in the following order-. This can be followed for all logarithmic expressions, given that the base is same for both the terms.

2. What is the division rule in logarithms?

If we have two or more logarithmic expressions with a similar base, their difference is equal to the logarithm of the ratio of both terms. For example-

3. For what values of the base, the logarithm is defined for an expression?

The logarithm for any number is defined for base>0 and *b* ≠ 1.

4. What is the value of the log of 1 over any base?

The value of log 1 is always zero over any base as X0=1, for any values of X.

5. Define the behavior of the logarithm function for different values of the base?

The logarithm function is a monotonously increasing function for base>1, and for 0<base<1, the logarithm is a monotonously decreasing function.

## Key takeaways

In this blog, we discussed the important concepts and formulas on logarithms, which will come in handy to prepare for the aptitude round of technical interviews. You can also consider our Aptitude Course to give your career an edge over others.