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# How do you solve for w in $\log w=\dfrac{1}{2}\log x+\log y$?

Last updated date: 10th Aug 2024
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Hint:First we transpose log (y) to the other side of the equation and then apply the quotient property of logarithm. Later we applied the power property of logarithm and then we solved the equation for the value of ‘w’ in a way we solved the general equations.

Formula used:
The quotient property of logarithm which states that${{\log }_{b}}m-{{\log }_{b}}n={{\log }_{b}}\dfrac{m}{n}$.
The power property of logarithm, i.e. $n\log a=\log {{\left( a \right)}^{n}}$.
The one-to-one property of logarithm, i.e. If$\log a=\log b$, then a = b

Complete step by step solution:
We have given that,
$\Rightarrow \log w=\dfrac{1}{2}\log x+\log y$
Transposing log(y) on the other side of the equation, we get
$\Rightarrow \log w-\log y=\dfrac{1}{2}\log x$
Using the quotient property of logarithm, i.e.
${{\log }_{b}}m-{{\log }_{b}}n={{\log }_{b}}\dfrac{m}{n}$
Applying the property, we get
$\Rightarrow \log \dfrac{w}{y}=\dfrac{1}{2}\log x$
Using the power property of logarithm, i.e.
$n\log a=\log {{\left( a \right)}^{n}}$
Applying the power property of log, we get
$\Rightarrow \log \dfrac{w}{y}=\log {{\left( x \right)}^{\dfrac{1}{2}}}$
Using the one-to-one property of logarithm, i.e.
If$\log a=\log b$, then a = b
Applying this property, we get
$\Rightarrow \dfrac{w}{y}={{\left( x \right)}^{\dfrac{1}{2}}}$
Solving for the value of w, we get
$\Rightarrow w={{\left( x \right)}^{\dfrac{1}{2}}}\times y=\sqrt{x}.y$
Therefore the value of w is equal to $\sqrt{x}y$or$\sqrt{x}.y$.