Question

How do you solve ${{10}^{x}}=30$ ? \[\]

Answer 1

Hint: We recall the definition of logarithm and identities of logarithm. We use the definition of ${{b}^{y}}=a\Rightarrow {{\log }_{b}}a=y$for $b=10,y=x,a=30$. We use the logarithmic identity of product ${{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n$ for $m=3,n=10$. We then use known logarithm value $\log 3=0.4771,\log 10=1$ to solve for $x$. \[\]

Complete step by step answer:
We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $a$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number $a$, which means if ${{b}^{y}}=a$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}a=y\]
Here $a,y,b$ are real numbers subjected to condition $a>0$ and $b>0,b\ne 1$.Here $a$ is called argument of logarithm. We know when base and argument are equal we have
\[{{\log }_{b}}b=1\]
We know the logarithmic identity involving power $m\ne 0$ where $m$ is real number as
\[m{{\log }_{b}}a={{\log }_{b}}{{a}^{m}}\]
We also know the logarithmic identity involving product as
\[{{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n\]
We are given the following equation to solve
\[{{10}^{x}}=30\]
We use definition of logarithm ${{b}^{y}}=a\Rightarrow {{\log }_{b}}a=y$ for base $b=10$, exponent $y=x$ and argument $a=30$
\[\begin{align}
  & {{\log }_{10}}30=x \\
 & \Rightarrow {{\log }_{10}}\left( 3\times 10 \right)=x \\
\end{align}\]
We use the logarithmic identity of product ${{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n$ for $m=3,n=10$ in the above step to have;
\[\Rightarrow {{\log }_{10}}3+\log {{ }_{10}}10=x\]
We use the identity for equal base and argument ${{\log }_{b}}b=1$ for $b=1$ and known value of ${{\log }_{10}}3=0.4771$ in the above step to have
\[\begin{align}
  & \Rightarrow 0.4771+1=x \\
 & \Rightarrow x=1.4771 \\
\end{align}\]

Note: We note that the if base is 10 we call the logarithm common logarithm and we can write the logarithm omitting the base 10. We can alternatively solve by taking common logarithm both sides of given equation ${{10}^{x}}=30$ to have;
\[\begin{align}
  & \log {{10}^{x}}=\log 30 \\
 & \Rightarrow \log {{10}^{x}}=\log \left( 3\times 10 \right) \\
\end{align}\]
We logarithmic identity involving power $m{{\log }_{b}}a={{\log }_{b}}{{a}^{m}}$ in the left hand side for $a=10,m=x$ and logarithmic identity product in right hand ${{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n$ for $m=3,m=10$ in the right hand side to have;
\[\Rightarrow x\log 10=\log 3+\log 10\]
We divide both sides of above equation by $\log 10$ and put the values $\log 3=0.4771,\log 10=1$ to have;
\[\begin{align}
  & x=\dfrac{\log 3+\log 10}{\log 10} \\
 & \Rightarrow x=\dfrac{0.4771+1}{1} \\
 & \Rightarrow x=1.4771 \\
\end{align}\]