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Question

Question: How do you solve \(\log x - \log 3 = 2\)?...

How do you solve logxlog3=2\log x - \log 3 = 2?

Explanation

Solution

Given a logarithmic equation. We have to find the solution of the expression. First, we will apply the basic rules of logarithms on the left hand side of the equation. Then, write the left hand side of the equation in a single term. Then, we will write the logarithmic expression in exponential form using the exponential rule of logarithms.
Formula used:
The logarithmic expression can be written in exponential form by:
If logbx=y{\log _b}x = y , then by=x{b^y} = x
The quotient rule of logarithms is given by:
logbxlogby=logbxy{\log _b}x - {\log _b}y = {\log _b}\dfrac{x}{y}

Complete step by step solution:
We are given the expression, logxlog3=2\log x - \log 3 = 2. First, we will apply the quotient rule of logarithms to the left hand side of the expression.
logx3=2\Rightarrow \log \dfrac{x}{3} = 2
Then, we will apply the exponential rule of logarithms to both sides of expression assuming the base 1010.
log10x3=2\Rightarrow {\log _{10}}\dfrac{x}{3} = 2
102=x3\Rightarrow {10^2} = \dfrac{x}{3}
On simplifying the expression, we get:
3×100=x\Rightarrow 3 \times 100 = x
x=300\Rightarrow x = 300
Hence, the solution of the expression x=300x = 300
Additional information: We are given the logarithmic expression. In the logarithmic expression, the quotient property of logarithms states that the difference of two logarithmic terms must be written as a log of division of two terms. Also the logarithmic expression can be represented in exponential form by applying the formula by=x{b^y} = x where b is the base of the logarithm, y is the constant term on right hand side of the expression and x is the logarithmic function given.

Note: A logarithm is a mathematical operation that determines how many times a certain number, called the base, is multiplied by itself to reach another number, we know that logarithm is the power to which a number must be raised in order to get some other number, and the base unit is the number being raised to a power. In these types of questions, we use logarithmic properties and formulas, and some of useful formulas are:
logaxy=logax+logay{\log _a}xy = {\log _a}x + {\log _a}y,
logxlogy=log(xy)\log x - \log y = \log \left( {\dfrac{x}{y}} \right)
logaxn=nlogax{\log _a}{x^n} = n{\log _a}x,
logab=logeblogea{\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}},
log1ab=logab{\log _{\dfrac{1}{a}}}b = - {\log _a}b,
logaa=1{\log _a}a = 1,
logaxb=1xlogab{\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b.