# Operations on Exponential Terms

Operations on Exponential Terms

Whenever we multiply two exponential terms with the same base, the exponents add. We talked about an intuitive explanation of this fact in the case of integer powers – the contributions from the different terms adding up. This property of exponential terms holds in general: $${b^p} \times {b^q} = {b^{\left( {p + q} \right)}}$$.

Similarly, upon dividing any two exponential terms with the same base, the exponents subtract: $${b^p} \div {b^q} = {b^{\left( {p - q} \right)}}$$.

Note that addition and subtraction of exponential terms does not lead to any similar combination of terms. For example, consider the expression $${2^7} + {2^{\frac{3}{2}}}$$. We cannot combine these two terms in any way. The most we can do is take out as the common factor one of the two terms:

${2^7} + {2^{\frac{3}{2}}} = {2^{\frac{3}{2}}}\left( {{2^{\left( {7 - \frac{3}{2}} \right)}} + 1} \right) = {2^{\frac{3}{2}}}\left( {{2^{\frac{{11}}{2}}} + 1} \right)$

What happens if we raise an exponential term to another power? The exponents multiply: $${\left( {{b^p}} \right)^q} = {b^{pq}}$$  Let’s try to understand this in the case where q is a positive integer:

\begin{align}&{\left( {{b^p}} \right)^q} = \overbrace {{b^p} \times {b^p} \times {b^p} \times ... \times {b^p}}^{q\;{\rm{times}}}\\&\qquad = \overbrace {{b^{\left( {p + p + ... + p} \right)}}}^{q\;{\rm{times}}}\\&\qquad = {b^{pq}}\end{align}

This property will hold even if q is a non-integer rational number, or even an irrational number. Try to understand how.

Using this property of multiplication of exponents, we can modify the base in any exponential term. For example,

\begin{align}&{3^4} = {\left( {{3^2}} \right)^2} = {9^2}, \quad {4^{10}} = {\left( {{2^2}} \right)^{10}} = {2^{20}}\\&{3^{2\pi }} = {\left( {{3^2}} \right)^\pi } = {9^\pi }\end{align}

Example 1: Find the value of x which will satisfy

(a) $${2^{x + 1}} = {4^{x + 7}}$$

(b) $${2^{2x - 1}} = {4^{x + 1}} \times {64^{ - 6x - 8}}$$

Solution: (a) We have

${2^{x + 1}} = {\left( {{2^2}} \right)^{x + 7}} = {2^{2x + 14}}$

Now, the exponents on both the sides can be compared:

$x + 1 = 2x + 14\,\,\, \Rightarrow x = - 13$

(b) We have

\begin{align}&{2^{2x - 1}} = {\left( {{2^2}} \right)^{x + 1}} \times {\left( {{2^6}} \right)^{ - 6x - 8}} = {2^{2x + 2}} \times {2^{ - 36x - 48}}\\&\qquad = {2^{\left( {2x + 2} \right) + \left( { - 36x - 48} \right)}} = {2^{ - 34x - 46}}\end{align}

Now, we can compare the exponents on both sides:

\begin{align}&2x - 1 = - 34x - 46\\ &\Rightarrow 36x = - 45\,\,\, \Rightarrow x = - \frac{{45}}{{36}} = - \frac{5}{4}\end{align}

Example 2: Consider the following relation:

$${\left( {{{\left( {{{\left( {{3^{ - 2}}} \right)}^{ - 3}}} \right)}^{ - 4}}} \right)^{\frac{1}{{12}}}} = {3^n}$$

What is the value of n?

Solution: We have

${\left( {{3^{ - 2}}} \right)^{ - 3}} = {3^6},\,\,\,{\left( {{3^6}} \right)^{ - 4}} = {3^{ - 24}},\,\,\,{\left( {{3^{ - 24}}} \right)^{\frac{1}{{12}}}} = {3^{ - 2}}$

Thus, $$n = - 2$$.

Example 3: Suppose that x and y are positive integers such that x is smaller, and

\begin{align}&(1)\quad{x^{\frac{1}{x}}} = {y^{\frac{1}{y}}}\\&(2)\quad{x^y} + {y^x} = 32\\&(3)\quad xy = 8\end{align}

Using these relations, deduce the value of $${x^{\frac{1}{x}}}$$.

Solution: We suppose that

\begin{align}&{x^{\frac{1}{x}}} = {y^{\frac{1}{y}}} = k\,\,\, \Rightarrow x = {k^x},\;y = {k^y}\\& \Rightarrow \left\{ \begin{array}{l}{x^y} = {\left( {{k^x}} \right)^y} = {k^{xy}}\\{y^x} = {\left( {{k^y}} \right)^x} = {k^{xy}}\end{array} \right.\,\,\\& \Rightarrow {x^y} + {y^x} = 2{k^{xy}} = 32\,\,\, \Rightarrow {k^{xy}} = 16 = {2^4}\end{align}

Since $$xy = 8$$, we have

${k^8} = {2^4} = {\left( {\sqrt 2 } \right)^8}\,\,\, \Rightarrow \,\,\,k = \sqrt 2$

Thus, $${x^{\frac{1}{x}}} = \sqrt 2$$.

Exponents and Logarithms
Grade 9 | Questions Set 1
Exponents and Logarithms
Exponents and Logarithms
Grade 9 | Questions Set 2
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More Important Topics
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More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus