Question

Solve the following equation for $x$ :
${{\log }_{9}}x=2.5$

Answer 1

In this problem we need to solve the given equation that means we have to calculate the value of $x$ which satisfies the given equation. We can observe that the given equation is logarithmic equation, so we will convert it into exponential form by using the logarithmic formula ${{\log }_{a}}x=b\Leftrightarrow {{a}^{b}}=x$ . After converting the given equation in exponential form we will write the value $2.5$ in fractional form as $\dfrac{5}{2}$ and the value $9$ in exponential form as ${{3}^{2}}$ . Now we will simplify the equation by using the exponential formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ and simplify the equation to get the required result.

Complete step-by-step solution:
Given equation is
${{\log }_{9}}x=2.5$.
Applying the logarithmic formula ${{\log }_{a}}x=b\Leftrightarrow {{a}^{b}}=x$ in the above equation, then we will get the exponential form as
${{9}^{2.5}}=x$
Isolating the variables in the above equation, then we will get
$x={{9}^{2.5}}$
We are writing the value $2.5$ in fractional form as $\dfrac{5}{2}$ and the value $9$ in exponential form as ${{3}^{2}}$ in the above equation, then we will have
$x={{\left( {{3}^{2}} \right)}^{\dfrac{5}{2}}}$
Applying the exponential formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ in the above equation, then we will get
$\begin{aligned} & x={{3}^{2\times \dfrac{5}{2}}} \\\ & \Rightarrow x={{3}^{5}} \\\ \end{aligned}$
Using the exponential formula ${{a}^{n}}=a\times a\times a\times a\times a.....\text{ n times}$ in the above equation, then we will have
$\begin{aligned} & x=3\times 3\times 3\times 3\times 3 \\\ & \Rightarrow x=243 \\\ \end{aligned}$
Hence the solution of the given equation ${{\log }_{9}}x=2.5$ is $x=243$ .

Note: In this problem we have used a couple of exponential and logarithmic formulas to get the result. Some of the exponential formulas which are useful in solving similar problems are given below
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ ,
${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$ .