# EXPONENTS AND LOGARITHMS - SOLVING FOR EXPONENT VARIABLE

Laws of Exponents before Exponents and Logarithms
if you have not already done so.

It is a prerequisite here.

There, we stated the 7 laws of indices
and the two Rules used for solving problems.

Here, we provide many Solved Examples

These problems deal with solving equations
for the variable which is an exponent.

We solve with the help of the knowledge
of Exponents. We give an example where
such solving requires the knowledge of
Logarithms and provide link to study them.

### Solved Example 1 of Exponents and Logarithms

If (-4)3 x (-4)7 = (-4)2a, find a.

Solution to Example 1 of Exponents and Logarithms:

By data,
(-4)3 x (-4)7 = (-4)2a
We know am x an = am + n.
Applying this Law to the L.H.S., we get
(-4)3 + 7 = (-4)2a
⇒ (-4)10 = (-4)2a
Since the bases are equal, the exponents should be equal.
∴ 10 = 2a. ⇒ 10⁄2 = a ⇒ 5 = a
Thus a = 5. Ans.

### Solved Example 2 of Exponents and Logarithms

If (2⁄3)4 x (2⁄3)-4 = (2⁄3)3a, find a.

Solution to Example 2 of Exponents and Logarithms:

By data,
(2⁄3)4 x (2⁄3)-4 = (2⁄3)3a.
We know am x an = am + n
and this law can be applied even to the negative integers.
Applying this Law, we get
(2⁄3)4 + (-4) = (2⁄3)3a.
⇒ (2⁄3)0 = (2⁄3)3a.
Since the bases are equal, the exponents should be equal.
∴ 0 = 3a. ⇒ 0⁄3 = a ⇒ 0 = a
Thus a = 0. Ans.

### Solved Example 3 of Exponents and Logarithms

If (5⁄2)2 x (5⁄2)a + 5 = (5⁄2)8, find a.

Solution to Example 3 of Exponents and Logarithms:

By data,
(5⁄2)2 x (5⁄2)a + 5 = (5⁄2)8
We know am x an = am + n
Applying this Law, we get
(5⁄2)2 + a + 5 = (5⁄2)8
⇒ (5⁄2)a + 7 = (5⁄2)8
Since the bases are equal, the exponents should be equal.
a + 7 = 8. ⇒ a = 8 - 7 = 1.
Thus a = 1. Ans.

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### Solved Example 4 of Exponents and Logarithms

Solved Example 4 of Exponents and Logarithms :

Find a so that 3a = 276.

Solution to Example 4 of Exponents and Logarithms:

By data,
3a = 276
We know 27 = 3 x 9 = 3 x 3 x 3 = 33
R.H.S. = 276 = (33)6
We know (am)n = amn
Applying this Law, we get
R.H.S. = 33 x 6 = 318
L.H.S. = 3a.
So, we have 3a = 318
Since the bases are equal, the exponents should be equal.
a = 18. Ans.

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### Solved Example 5 of Exponents and Logarithms

Solved Example 5 of Exponents and Logarithms :

Find m so that 4 x 8m = 25

Solution to Example 5 of Exponents and Logarithms:

By data,
4 x 8m = 25
We know 4 = 2 x 2 = 22 and 8 = 2 x 2 x2 = 23
L.H.S. = 22 x (23)m
We know (am)n = amn
Applying this Law, we get
L.H.S. = 22 x 23m
We know am x an = am + n
Applying this Law, we get
L.H.S. = 22 + 3m
R.H.S. = 25
So, we have 22 + 3m = 25
Since the bases are equal, the exponents should be equal.
∴ 2 + 3m = 5 ⇒ 3m = 5 - 2 = 3 ⇒ m = 3⁄3 = 1.
m = 1. Ans.

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### Solved Example 6 of Exponents and Logarithms

Solved Example 6 of Exponents and Logarithms :

Determine x so that (5⁄4)-5 x (5⁄4)-10 = (5⁄4)3x

Solution to Example 6 of Exponents and Logarithms:

By data,
(5⁄4)-5 x (5⁄4)-10 = (5⁄4)3x
We know am x an = am + n
and this law can be applied even to the negative integers.
Applying this Law, we get
L.H.S. = (5⁄4)-5 + (-10) = (5⁄4)-15
R.H.S. = (5⁄4)3x
So, we have (5⁄4)-15 = (5⁄4)3x
Since the bases are equal, the exponents should be equal.
∴ -15 = 3x ⇒ -15⁄3 = x. ⇒ -5 = x.
x = -5. Ans.

### Need of Logarithms

Let me put a question.
If 2x = 8, What is x ?
You are ready to answer as: Since 2x = 8 = 2 x 2 x 2 = 23,
and as the bases are same, the exponents should be equal.∴ x = 2.
Now, instead of 8, if I give you 7 on the R.H.S.,
i.e. If 2x = 7, What is x ? To answer this question, we have to go to

LOGARITHMS

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