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Question

Question: How do you solve \(\log \left( x \right) = - 0.123\)?...

How do you solve log(x)=0.123\log \left( x \right) = - 0.123?

Explanation

Solution

In order to solve this we need to know the property of the logarithm and there are many such properties but we need to use them according to our need. We will use the property which says that:
If logab=c{\log _a}b = c then we can say that ac=b{a^c} = b.

Complete step by step solution:
Here we are given to solve the logarithmic function which is given as log(x)=0.123\log \left( x \right) = - 0.123.
If we are given logab=c{\log _a}b = c then we must know that here aa is the base of the logarithm.
Here we must know that if we are given lnx\ln x then its base is taken as e=2.718e = 2.718 and if it is simply given as logx\log x then the base is actually taken as 1010 over here. Hence we must be clear with this point.
Now we need to solve for log(x)=0.123\log \left( x \right) = - 0.123 which means we need to find the value of xx.
We know the property of log\log which says that:
If logab=c{\log _a}b = c then we can say that ac=b{a^c} = b.
Now we know that in the above problem we have to take the base at 1010.
So we can compare logab=c{\log _a}b = c with the given logarithmic function which is log(x)=0.123\log \left( x \right) = - 0.123.
So we will get:
a=10 b=x c=0.123  a = 10 \\\ b = x \\\ c = - 0.123 \\\
So we know that if logab=c{\log _a}b = c then we can say that ac=b{a^c} = b
So we can write 100.123=x{10^{ - 0.123}} = x
x=100.123x = {10^{ - 0.123}}
Now we can solve this with the calculator and get the value as:
x=100.123=0.7534x = {10^{ - 0.123}} = 0.7534.

Note: Whenever the student is given to solve the problems which contain logarithmic function, he must know the properties of the log and also when to use which formula. Hence this is very necessary and it comes by practice. The properties of log\log are like:
log(ab)=loga+logb log(ab)=logalogb  \log (ab) = \log a + \log b \\\ \log \left( {\dfrac{a}{b}} \right) = \log a - \log b \\\
logmn=nlogm\log {m^n} = n\log m.