Question

Solve the following equation and find the value of \[x\] , \[{{\log }_{2}}\left( {{25}^{x+3}}-1 \right)=2+{{\log }_{2}}\left( {{5}^{x+3}}+1 \right)\].

Answer 1

Hint: First of all, use the property \[{{\log }_{a}}a=1\] for the constant term 2 in the given equation and modify the given equation. Now, use the property of logarithm, \[b{{\log }_{c}}a={{\log }_{c}}{{a}^{b}}\] to simplify it further. Use the formula, \[{{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n\] to convert the RHS of the modified equation into a single logarithmic term and then apply antilog to remove the log. Assume \[p={{5}^{x+3}}\] and then solve the quadratic equation to get the value of p. Ignore the negative value of p because p cannot be negative. Now, with the help of p, get the value of x.

Complete step-by-step solution
According to the question, we are given an equation and we have to find the value of \[x\].
The given equation is \[{{\log }_{2}}\left( {{25}^{x+3}}-1 \right)=2+{{\log }_{2}}\left( {{5}^{x+3}}+1 \right)\] ………………………………………..(1)
We know the property of logarithm, \[{{\log }_{a}}a=1\] ……………………………………(2)
Now, on putting \[a=2\] in equation (2), we get
\[\Rightarrow {{\log }_{2}}2=1\] …………………………………………………(3)
From equation (1) and equation (3), we get
\[\Rightarrow {{\log }_{2}}\left( {{25}^{x+3}}-1 \right)=2\times 1+{{\log }_{2}}\left( {{5}^{x+3}}+1 \right)\]
\[\Rightarrow {{\log }_{2}}\left( {{25}^{x+3}}-1 \right)=2\times {{\log }_{2}}2+{{\log }_{2}}\left( {{5}^{x+3}}+1 \right)\] ……………………………………………(4)
We also know the property of logarithm, \[b{{\log }_{c}}a={{\log }_{c}}{{a}^{b}}\] ……………………………………………(5)
Now, using the property shown in equation (5) and simplifying equation (4), we get
\[\Rightarrow {{\log }_{2}}\left( {{25}^{x+3}}-1 \right)={{\log }_{2}}{{2}^{2}}+{{\log }_{2}}\left( {{5}^{x+3}}+1 \right)\] …………………………………………(6)
We have to further simplify the above equation
We also know the property of logarithm, \[{{\log }_{a}}mn={{\log }_{a}}m+{{\log }_{a}}n\] …………………………………………..(7)
Now, from equation (6) and equation (7), we get
 \[\Rightarrow {{\log }_{2}}\left( {{25}^{x+3}}-1 \right)={{\log }_{2}}{{2}^{2}}\left( {{5}^{x+3}}+1 \right)\]
\[\Rightarrow {{\log }_{2}}\left( {{25}^{x+3}}-1 \right)={{\log }_{2}}4\left( {{5}^{x+3}}+1 \right)\] ……………………………………………..(8)
On applying antilog in LHS and RHS of the above equation, we get
 \[\Rightarrow \left( {{25}^{x+3}}-1 \right)=4\left( {{5}^{x+3}}+1 \right)\]
\[\Rightarrow {{5}^{2\left( x+3 \right)}}-1=4\times {{5}^{x+3}}+4\] …………………………………………….(9)
Let us assume \[p={{5}^{x+3}}\] ……………………………………………(10)
From equation (9) and equation (10), we get
 \[\begin{align}
  & \Rightarrow {{p}^{2}}-1=4\times p+4 \\
 & \Rightarrow {{p}^{2}}-4p-1-4=0 \\
 & \Rightarrow {{p}^{2}}-4p-5=0 \\
 & \Rightarrow {{p}^{2}}-5p+p-5=0 \\
 & \Rightarrow p\left( p-5 \right)+1\left( p-5 \right)=0 \\
 & \Rightarrow \left( p+1 \right)\left( p-5 \right)=0 \\
\end{align}\]
So, \[p=-1\] and \[p=5\] .
Here, we have two values of p. The value \[p=-1\] is not possible because \[p={{5}^{x+3}}\] which cannot be a negative number. So, we have to ignore \[p=-1\] .
Now, let us take \[p=5\] …………………………………………(11)
From equation (10) and equation (11), we get
\[\Rightarrow 5={{5}^{x+3}}\]
\[\Rightarrow {{5}^{1}}={{5}^{x+3}}\] …………………………………………..(12)
On comparing the exponents of LHS and RHS of the above equation, we get
\[\begin{align}
  & \Rightarrow 1=x+3 \\
 & \Rightarrow 1-3=x \\
 & \Rightarrow -2=x \\
\end{align}\]
Therefore, the value of x is equal to -2, \[x=-2\].

Note: Whenever this type of question appears where some part of LHS and RHS is in logarithm along with some constant terms. Always approach this type of question by converting the LHS into one logarithmic term. Similarly, convert the RHS into one logarithmic term. Doing this will make our LHS and RHS easy to be compared and solved.