Question

How do you solve: ${3^{10}} = {9^x}$?

Answer 1

Hint: The problem deals with comparing the powers or indices of two numbers using the basic laws of exponents. Since the two numbers whose powers are to be equated are not the same, we have to first make the bases of exponents the same before comparing the exponents. For making the bases the same, we use basic exponent rules.

Complete step-by-step solution:
Making use of laws of exponents to make the bases same because exponents can be compared only when the bases are equal. We know that $9 = {3^2}$ .
So, we have, ${3^{10}} = {9^x}$
Now, we substitute the value of $9$ as ${3^2}$.
\[ \Rightarrow \]${3^{10}} = {\left( {{3^2}} \right)^x}$\[\]
Now, using law of exponent ${\left( {{a^n}} \right)^m} = {\left( a \right)^{nm}}$, we get
\[ \Rightarrow \]${3^{10}} = {3^{2 \times x}}$
Simplifying the expression, we get,
\[ \Rightarrow \] ${3^{10}} = {3^{2x}}$
Now we have the same bases on both sides, so now we can equate exponents or powers of both sides of the equation. Comparing the exponents,
\[ \Rightarrow \]$10 = 2x$
Using transposition rule and dividing both sides of equation by $2$,
\[ \Rightarrow \]$x = \dfrac{{10}}{2} = 5$
So, we get $x = 5$ on solving the given exponential equation by equating exponents after making the bases the same using laws of exponents.

Note: We can also solve the given exponential equation by use of logarithms by taking \[\log \] to the base \[10\]on both sides. These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be moulded according to our convenience while solving the problem. Also note that cube-root, square-root are fractions with 1 as numerator and respective root in denominator.