Question

Find the value of \[x\] if \[{\log _{11}}\left( {2x - 1} \right) = 1 - {\log _{11}}\left( {x + 4} \right)\].

Answer 1

Hint: To find the value of \[x\], we will first rearrange it to write in the form \[{\log _{11}}\left( {2x - 1} \right) + {\log _{11}}\left( {x + 4} \right) = 1\]. Then we will use the product rule in LHS and the power rule in the RHS to rewrite it. Then we will write the equation without \[\log \] and simplify it to find \[x\].

Complete step by step answer:
Given, \[{\log _{11}}\left( {2x - 1} \right) = 1 - {\log _{11}}\left( {x + 4} \right)\].
Taking \[{\log _{11}}\left( {x + 4} \right)\] to LHS, we get
\[ \Rightarrow {\log _{11}}\left( {2x - 1} \right) + {\log _{11}}\left( {x + 4} \right) = 1\]

As we know from the product rule that \[{\log _b}(mn) = {\log _b}(m) + {\log _b}(n)\]. Using this, we get
\[ \Rightarrow {\log _{11}}\left[ {\left( {2x - 1} \right)\left( {x + 4} \right)} \right] = 1\]
Now, from the power rule we know that \[{\log _b}\left( {{m^p}} \right) = p{\log _b}m\]. Using this, we can write \[1\] as \[{\log _{11}}\left( {{{11}^1}} \right)\].
\[ \Rightarrow {\log _{11}}\left[ {\left( {2x - 1} \right)\left( {x + 4} \right)} \right] = {\log _{11}}\left( {{{11}^1}} \right)\]
\[ \Rightarrow {\log _{11}}\left[ {\left( {2x - 1} \right)\left( {x + 4} \right)} \right] = {\log _{11}}\left( {11} \right)\]

Both sides of the equation contain \[\log \] terms with the same base. The equation is valid without \[\log \] also. So, we can write
\[ \Rightarrow \left( {2x - 1} \right)\left( {x + 4} \right) = 11\]
Multiplying the terms, we get
\[ \Rightarrow 2{x^2} + 8x - x - 4 = 11\]
\[ \Rightarrow 2{x^2} + 7x - 4 = 11\]
Subtracting \[11\] from both the sides, we get
\[ \Rightarrow 2{x^2} + 7x - 15 = 0\]

Splitting the middle term, we get
\[ \Rightarrow 2{x^2} + 10x - 3x - 15 = 0\]
Taking common, we get
\[ \Rightarrow 2x\left( {x + 5} \right) - 3\left( {x + 5} \right) = 0\]
\[ \Rightarrow \left( {x + 5} \right)\left( {2x - 3} \right) = 0\]
On solving, we get
\[ \Rightarrow x = - 5\] or \[x = \dfrac{3}{2}\]
As we know that \[{\log _b}a\] is defined for \[a,b > 0\] and \[b \ne 1\]. Therefore, \[x \ne - 5\].
So, \[x = \dfrac{3}{2}\].

Therefore, the value of \[x\] is \[\dfrac{3}{2}\].

Note: It is very important to note that \[\log x\] denotes that the base is \[10\] and \[\ln x\] denotes that base is \[e\]. Also, \[{\log _b}x\] is only defined when \[b\] and \[x\] are two positive real numbers and b is not equal to \[1\]. Logarithm base \[10\] i.e., \[b = 10\] is called decimal or common logarithm, logarithm base \[e\] is called natural logarithm and binary logarithm uses base \[2\].