Question
Question: How do you solve: \(3\log 2x=4\)?...
How do you solve: 3log2x=4?
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 then the relation between a,b&c is a=bc. 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=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=34
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=34
If not given then we have to take the base of the above logarithm as 10 so rewriting the above equation we get,
log102x=34
Now, we are going to use the following logarithm property to solve the above equation as follows:
logba=c
Then the relation between x, y and z is as follows:
a=bc
Substituting a=2x,b=10,c=34 in the above equation we get,
2x=(10)34
Dividing 2 on both the sides of the above equation we get,
22x=21(10)34
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=21(10)34
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 x which we are getting in the above solution is 21(10)34 and the equation given above is as follows:
3log2x=4
Substituting x=21(10)34 in the above equation we get,
3log221(10)34=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)34=4 ………(1)
Now, we are going to use the following property of logarithm which states that:
logxa=alogx
Substituting x=10 and a=34 in the above equation and we get,
log1034=34log10
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,
log1034=34log1010⇒log1034=34(1)⇒log1034=34
Using above relation in eq. (1) we get,
3log(10)34=4⇒3(34)=4⇒4=4
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.