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, \({x^{\frac{1}{x}}} = \sqrt 2 \).