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Question

Question: How do you solve: \(3\log 2x=4\)?...

How do you solve: 3log2x=43\log 2x=4?

Explanation

Solution

To solve the above logarithmic equation we first of all going to divide both the sides by 3 then we are going to use the following logarithm property which states that: logba=c{{\log }_{b}}a=c then the relation between a,b&ca,b\And c is a=bca={{b}^{c}}. And in this way, we can get the value of x.

Complete step-by-step solution:
The logarithmic expression given in the above problem is as follows:
3log2x=43\log 2x=4
We are asked to find the solution to the above equation which we are going to solve by dividing 3 on both the sides of the above equation and we get,
33log2x=43\dfrac{3}{3}\log 2x=\dfrac{4}{3}
In the above equation, 3 will get cancelled out from the numerator and the denominator of the L.H.S of the above equation and we get,
log2x=43\log 2x=\dfrac{4}{3}
If not given then we have to take the base of the above logarithm as 10 so rewriting the above equation we get,
log102x=43{{\log }_{10}}2x=\dfrac{4}{3}
Now, we are going to use the following logarithm property to solve the above equation as follows:
logba=c{{\log }_{b}}a=c
Then the relation between x, y and z is as follows:
a=bca={{b}^{c}}
Substituting a=2x,b=10,c=43a=2x,b=10,c=\dfrac{4}{3} in the above equation we get,
2x=(10)432x={{\left( 10 \right)}^{\dfrac{4}{3}}}
Dividing 2 on both the sides of the above equation we get,
2x2=12(10)43\dfrac{2x}{2}=\dfrac{1}{2}{{\left( 10 \right)}^{\dfrac{4}{3}}}
In the above equation, in the L.H.S of the above equation, 2 will get cancelled out from the numerator and the denominator and we get,
x=12(10)43x=\dfrac{1}{2}{{\left( 10 \right)}^{\dfrac{4}{3}}}
Hence, we have found the solution of the given equation in x.

Note: We can check whether the given value of x is correct or not by substituting that value of x in the above equation and see if that value is satisfying the given equation or not.
The value of xx which we are getting in the above solution is 12(10)43\dfrac{1}{2}{{\left( 10 \right)}^{\dfrac{4}{3}}} and the equation given above is as follows:
3log2x=43\log 2x=4
Substituting x=12(10)43x=\dfrac{1}{2}{{\left( 10 \right)}^{\dfrac{4}{3}}} in the above equation we get,
3log2(12(10)43)=43\log 2\left( \dfrac{1}{2}{{\left( 10 \right)}^{\dfrac{4}{3}}} \right)=4
In the L.H.S of the above equation, in the logarithm, 2 will get cancelled out from the numerator and the denominator and we get,
3log((10)43)=43\log \left( {{\left( 10 \right)}^{\dfrac{4}{3}}} \right)=4 ………(1)
Now, we are going to use the following property of logarithm which states that:
logxa=alogx\log {{x}^{a}}=a\log x
Substituting x=10x=10 and a=43a=\dfrac{4}{3} in the above equation and we get,
log1043=43log10\log {{10}^{\dfrac{4}{3}}}=\dfrac{4}{3}\log 10
We know that if not given then base of the logarithm is taken as 10 so taking the base as 10 in the above logarithm we get,
log1043=43log1010 log1043=43(1) log1043=43 \begin{aligned} & \log {{10}^{\dfrac{4}{3}}}=\dfrac{4}{3}{{\log }_{10}}10 \\\ & \Rightarrow \log {{10}^{\dfrac{4}{3}}}=\dfrac{4}{3}\left( 1 \right) \\\ & \Rightarrow \log {{10}^{\dfrac{4}{3}}}=\dfrac{4}{3} \\\ \end{aligned}
Using above relation in eq. (1) we get,
3log((10)43)=4 3(43)=4 4=4 \begin{aligned} & 3\log \left( {{\left( 10 \right)}^{\dfrac{4}{3}}} \right)=4 \\\ & \Rightarrow 3\left( \dfrac{4}{3} \right)=4 \\\ & \Rightarrow 4=4 \\\ \end{aligned}
In the above equation, L.H.S = R.H.S so the value of x which we have found in the above solution is correct.