EXPONENTS AND LOGARITHMS -
SOLVING FOR
EXPONENT VARIABLE

Please study
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
and Exercises with answers.

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 a^m x aⁿ = a^{m + 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 a^m x aⁿ = a^{m + 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 a^m x aⁿ = a^{m + 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 3^a = 27⁶.

Solution to Example 4 of Exponents and Logarithms:

By data,
3^a = 27⁶
We know 27 = 3 x 9 = 3 x 3 x 3 = 3³
R.H.S. = 27⁶ = (3³)⁶
We know (a^m)ⁿ = a^mn
Applying this Law, we get
R.H.S. = 3^{3 x 6} = 3¹⁸
L.H.S. = 3^a.
So, we have 3^a = 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 8^m = 2⁵

Solution to Example 5 of Exponents and Logarithms:

By data,
4 x 8^m = 2⁵
We know 4 = 2 x 2 = 2² and 8 = 2 x 2 x2 = 2³
L.H.S. = 2² x (2³)^m
We know (a^m)ⁿ = a^mn
Applying this Law, we get
L.H.S. = 2² x 2^3m
We know a^m x aⁿ = a^{m + n}
Applying this Law, we get
L.H.S. = 2^{2 + 3m}
R.H.S. = 2⁵
So, we have 2^{2 + 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 a^m x aⁿ = a^{m + 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 2^x = 8, What is x ?
You are ready to answer as: Since 2^x = 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 2^x = 7, What is x ? To answer this question, we have to go to

LOGARITHMS

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