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