Question

How do you solve: \[\log (x + 9) - \log (x) = 1?\]

Answer 1

Hint: Here we have to solve the given term by using logarithm formulas and then we will compare both the sides to find the value of \[x\]. Finally we get the required answer.

Formula used: We will use following logarithm formulas:
\[1.\] For multiplication term inside logarithm-
\[\log (MN) = \log (M) + \log (N)\]
\[2.\] For division term inside the logarithm-
\[\log \left( {\dfrac{M}{N}} \right) = \log (M) - \log (N)\]
\[3.\] For the same term in base and value-
\[{\log _a}a = 1\], where ‘\[a\]’can be any variable or constant term.
\[4.\]Cross-multiplication of base to the terms in the other side of the equation-
\[{\log _x}y = a\]
\[ \Rightarrow {x^a} = y\]

Complete step-by-step solution:
So, we can write the given equation as follows:
\[\log (x + 9) - \log (x) = 1\]
\[ \Rightarrow \log \left( {\dfrac{{x + 9}}{x}} \right) = 1\] (After using the above formula)
We know that \['10'\]is the general base value of any logarithm.
So, we can rewrite the above equation as following:
\[\log (x + 9) - \log (x) = 1\]
\[ \Rightarrow {\log _{10}}\left( {\dfrac{{x + 9}}{x}} \right) = 1\]
But we can write R.H.S as the same base of \['10'\] in base as well as value of logarithm.
So, we can write \[1\] as\[{\log _{10}}10\].
So, by substituting R.H.S with\[1 = {\log _{10}}10\], we get:
\[\log (x + 9) - \log (x) = 1\]
\[ \Rightarrow {\log _{10}}\left( {\dfrac{{x + 9}}{x}} \right) = {\log _{10}}10\]
So, it is clear that the base of the both part of the L.H.S and R.H.S are equal.
So, according to the rule of logarithm, the value of logarithm of an equation will be equal if their bases are equal.
So, by comparing the above equation, we can write the following equation:
\[ \Rightarrow \dfrac{{x + 9}}{x} = 10\]
Multiply the denominator term in L.H.S to the R.H.S, we get:
\[ \Rightarrow (x + 9) = 10x\].
Now, take variable term \[x\] to the other side of the equation, we get:
\[ \Rightarrow 9 = 10x - x\].
Now, by subtracting these, we get:
\[ \Rightarrow 9x = 9\].
Now, divide the R.H.S by \[9\], we get:
\[ \Rightarrow x = \dfrac{9}{9}\]
\[ \Rightarrow x = 1\]

\[\therefore \]The value of \[x\] is \[1\].

Note: Always remember that the value inside the logarithm should be positive, as \[\log ( - \cos \tan t)\] is undefined in mathematics.
Therefore, \[(x + 9) > 0\] as well as \[x > 0\].
It states that, \[x > - 9\].
As \[x > 0\] is the more restrictive condition of the equation, the domain of our function should be\[x > 0\].