Question

How do you solve \[{{5}^{{{x}^{2}}+2x}}=125\]?

Answer 1

Hint: In the given question, we have been asked to find the value of ‘x’ and it is given that \[{{5}^{{{x}^{2}}+2x}}=125\]. In order to solve the question, first we write the numbers in exponential form. After that we will equate the powers with each other, we will get the quadratic equation and solve in a way we solve the general quadratic equation. Later write the quadratic equation in a standard form and then find the roots of the given quadratic equation. We solve resultant quadratic equations in a way we solve general quadratic equations.

Complete step by step solution:
We have given that,
\[\Rightarrow {{5}^{{{x}^{2}}+2x}}=125\]
As we know that,
\[\Rightarrow 5\times 5\times 5={{5}^{3}}=125\]
Substituting \[125={{5}^{3}}\] in the given equation, we get
\[\Rightarrow {{5}^{{{x}^{2}}+2x}}={{5}^{3}}\]
Now both the sides base are equal i.e. 5.
Equate powers with each other, we get
\[\Rightarrow {{x}^{2}}+2x=3\]
Now solving for the value of ‘x’.
Rewrite the equation in a standard quadratic form, we get
\[\Rightarrow {{x}^{2}}+2x-3=0\]
Splitting the middle term, we get
\[\Rightarrow {{x}^{2}}+3x-1x-3=0\]
Taking out common factor by making the pairs, we get
\[\Rightarrow x\left( x+3 \right)-1\left( x+3 \right)=0\]
Taking out common factors, we get
\[\Rightarrow \left( x-1 \right)\left( x+3 \right)=0\]
Equating each common factors equals to 0, we get
\[\Rightarrow x-1=0\ and\ x+3=0\]
Solving for the value of ‘x’, we get
\[\Rightarrow x=1\ and\ x=-3\]
Therefore, the possible values of \[x\] are 1 and -3.

Note: While solving these types of questions, students need to remember the properties of exponent and powers. For converting the number into exponential form, we will need to find the prime factorization of that number and we will get our number in exponential form. Exponents are used to write the same number in multiplication form using a simple format.