Question

Express \[a\] in terms of \[b\] in the following if
1) \[\dfrac{1}{2}\log a + 5\log b = 1\]
2) \[2\log a - \dfrac{1}{3}\log b = 2\]

Answer 1

Hint:
We will find \[a\] in terms of \[b\] by manipulating the given linear equation using basic mathematical operations like addition, subtraction, multiplication and division. First, we will find \[\log a\] in terms of \[\log b\] and then we will find the antilog by raising the logarithm to the power of its base.
Formulas used: We will use the following formulas to solve the questions:
1. \[{x^{p + q}} = {x^p} \cdot {x^q}\]
2. \[{x^{p - q}} = \dfrac{{{x^p}}}{{{x^q}}}\]
3. \[{10^{{{\log }_{10}}x}} = {x^{{{\log }_{10}}10}}\]
4. \[{\log _z}z = 1\]

Complete step by step solution:
(i) First, let us subtract \[5\log b\] from both sides of the equation.
\[\begin{array}{l}\dfrac{1}{2}\log a + 5\log b - 5\log b = 1 - 5\log b\\ \Rightarrow\dfrac{1}{2}\log a = 1 - 5\log b\end{array}\]
Now, let’s multiply both sides of the equation by 2.
\[\begin{array}{l}2\left( {\dfrac{1}{2}\log a} \right) = 2\left( {1 - 5\log b} \right)\\ \Rightarrow\log a = 2 - 10\log b\end{array}\]
We have found \[\log a\] in terms of \[\log b\]. Now, we will find the antilog. As the base of the log is 10, let’s raise both sides to the power of 10.
\[{10^{{{\log }_{10}}a}} = {10^{2 - 10{{\log }_{10}}b}}\]
We will now apply the formula \[{x^{p - q}} = \dfrac{{{x^p}}}{{{x^q}}}\] to the R.H.S of the equation obtained.
Substituting 10 for \[x\] , 2 for \[p\] and \[10{\log _{10}}b\] for
\[q\], we get
\[\begin{array}{l}{10^{{{\log }_{10}}a}} = \dfrac{{{{10}^2}}}{{{{10}^{10{{\log }_{10}}b}}}}\\ \Rightarrow{10^{{{\log }_{10}}a}} = \dfrac{{100}}{{{{10}^{10{{\log }_{10}}b}}}}\end{array}\]
Now we will use the formula \[{10^{{{\log }_{10}}x}} = {x^{{{\log }_{10}}10}}\] and \[{\log _z}z = 1\] to get \[a\] in terms of \[b\].
Substituting \[a\] for \[x\] in the L.H.S, \[b\] for \[x\] in the R.H.S and 10 for \[z\], we get
\[\begin{array}{l}{a^{{{\log }_{10}}10}} = \dfrac{{100}}{{{b^{10{{\log }_{10}}10}}}}\\ \Rightarrow{a^1} = \dfrac{{100}}{{{b^{10 \cdot 1}}}}\\ \Rightarrow a = 100{b^{ - 10}}\end{array}\]

\[\therefore\] \[a\] can be expressed as \[100{b^{ - 10}}\].

(ii) First, let’s add \[\dfrac{1}{3}\log b\] on both sides of the equation.
\[\begin{array}{l}2\log a - \dfrac{1}{3}\log b + \dfrac{1}{3}\log b = 2 + \dfrac{1}{3}\log b\\ \Rightarrow2\log a = 2 + \dfrac{1}{3}\log b\end{array}\]
Dividing both sides by 2, we get
 \[\begin{array}{l}\dfrac{{2\log a}}{2} = \dfrac{{2 + \dfrac{1}{3}\log b}}{2}\\ \Rightarrow\log a = \dfrac{2}{2} + \dfrac{1}{{3 \cdot 2}}\log b\\ \Rightarrow\log a = 1 + \dfrac{1}{6}\log b\end{array}\]
We have found \[\log a\] in terms of \[\log b\].
Now, we will find the antilog. As the base of the log is 10, let’s raise both sides to the power of 10.
\[{10^{{{\log }_{10}}a}} = {10^{1 + \dfrac{1}{6}{{\log }_{10}}b}}\]
We will now apply the formula \[{x^{p - q}} = \dfrac{{{x^p}}}{{{x^q}}}\] to the R.H.S of the equation obtained.
Substituting 10 for \[x\] , 1 for \[p\] and \[\dfrac{1}{6}{\log _{10}}b\] for \[q\], we get
\[{10^{{{\log }_{10}}a}} = {10^1} \cdot {10^{\dfrac{1}{6}{{\log }_{10}}b}}\]
Now we will use the formula \[{10^{{{\log }_{10}}x}} = {x^{{{\log }_{10}}10}}\] and \[{\log _z}z = 1\] to get \[a\] in terms of \[b\].
Substituting \[a\] for \[x\] in the L.H.S, \[b\] for \[x\] in the R.H.S and 10 for \[z\], we get
\[\begin{array}{l}{10^{{{\log }_{10}}a}} = {10^1} \cdot {10^{\dfrac{1}{6}{{\log }_{10}}b}}\\ \Rightarrow{a^{{{\log }_{10}}10}} = 10 \cdot {b^{\dfrac{1}{6}{{\log }_{10}}10}}\\ \Rightarrow a = 10{b^{\dfrac{1}{6}}}\end{array}\]

\[\therefore\] \[a\] can be expressed as \[10{b^{\dfrac{1}{6}}}\].

Note:
It is important to know that when no base is mentioned in the case of a logarithm, the base is assumed to be 10. Similarly, if instead of \[\log a\] , \[\ln a\] is mentioned, we will take the base as \[e\]. If any other number is the base (say 7), we will raise both sides of the equation to the power of that number (7) to find the antilog.