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Question

Question: How do you solve \[{\log _{10}}(x - 6) = {\log _{10}}(2x + 1)\]?...

How do you solve log10(x6)=log10(2x+1){\log _{10}}(x - 6) = {\log _{10}}(2x + 1)?

Explanation

Solution

Hint : Here the question is related to the logarithmic functions. The log terms which are used in the question is common logarithm. Applying the antilog on both sides the log terms will get cancelled and hence we determine the value “x”. Hence, we obtain the required result for the given question.

Complete step-by-step answer :
In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number xx is the exponent to which another fixed number, the base bb, must be raised, to produce that number xx
In logarithmic functions we have two kinds: one is natural logarithm and other one is common logarithm. In natural logarithm the base value is “e” which is an exponent. In the common logarithm, base value is 10.
Now consider the given question
log10(x6)=log10(2x+1){\log _{10}}(x - 6) = {\log _{10}}(2x + 1)
On applying the antilog on both sides, we have
(x6)=(2x+1)\Rightarrow (x - 6) = (2x + 1)
Take x to the RHS and 1 to the LHS, on shifting the terms we have
61=2xx\Rightarrow - 6 - 1 = 2x - x
On simplifying we get
x=7\Rightarrow x = - 7
Hence, we have determined the value of x. If we substitute the value of x in the given question, we check whether the obtained answer is correct or not.
Consider the given question
log10(x6)=log10(2x+1){\log _{10}}(x - 6) = {\log _{10}}(2x + 1)
Substitute the value of x as -7, hence we get
log10(76)=log10(2(7)+1)\Rightarrow {\log _{10}}( - 7 - 6) = {\log _{10}}(2( - 7) + 1)
On simplifying we get
log10(13)=log10(14+1)\Rightarrow {\log _{10}}( - 13) = {\log _{10}}( - 14 + 1)
log10(13)=log10(13)\Rightarrow {\log _{10}}( - 13) = {\log _{10}}( - 13)
Hence verified.
So, the correct answer is “x = - 7”.

Note : While transforming or shifting terms the sign will change. So we should not take care of the sign. By substituting the value of x we can verify the given question. The logarithm and antilogarithm are the inverse of each other, so we can cancel each other.