Question

The equation \[{125^x} + {45^x} = 2{(27)^x}\] has
A.No solution
B.One solution
C.Two solution
D.More than two solutions

Answer 1

Hint: Here, we have to find the number of solutions by solving the given equation. First, we rewrite the given equation in terms of a square or multiple of a smaller number. Then we will solve the quadratic equation to arrive at a solution. An equation where the highest exponent of the variable is a square is a quadratic equation.

Formula used:
 We will use the following formulas:
1.The cube of difference of two numbers is given by the algebraic identity
$a^3-b^3$=$ (a - b)({a^2} + ab + {b^2})$
2.Exponential Formula: \[\left( {\dfrac{{{a^m}}}{{{b^m}}}} \right) = {\left( {\dfrac{a}{b}} \right)^m}\]

Complete step-by-step answer:
We are given with an equation \[{125^x} + {45^x} = 2{(27)^x}\]
Rewriting the given equation in the form of square and cube of integers, we have
\[ \Rightarrow {\left( {{5^3}} \right)^x} + {\left( {9 \times 5} \right)^x} = 2{(3 \times 9)^x}\]
 \[ \Rightarrow {\left( {{5^3}} \right)^x} + {\left( {{3^2} \times 5} \right)^x} = 2{(3 \times {3^2})^x}\]
\[ \Rightarrow {\left( {{5^x}} \right)^3} + {({3^x})^2} \times {5^x} = 2{({3^x})^3}\]
Now, taking \[{5^x} = a\] and \[{3^x} = b\], we get
 \[ \Rightarrow {a^3} + a{b^2} - 2{b^3} = 0\]
Rearranging the terms, we get
\[ \Rightarrow {a^3} - {b^3} + a{b^2} - {b^3} = 0\]
Now, by using the formula $a^3-b^3$=$ (a - b)({a^2} + ab + {b^2})$ and rewriting the equation, we have
\[ \Rightarrow (a - b)({a^2} + ab + {b^2}) + (a - b){b^2} = 0\]
Rewriting the equation again, we have
\[ \Rightarrow (a - b)({a^2} + ab + {b^2} + {b^2}) = 0\]
Adding the like terms, we have
\[ \Rightarrow (a - b)({a^2} + ab + 2{b^2}) = 0\]
Rewriting the equation, we have
\[ \Rightarrow a - b = 0\]
\[ \Rightarrow a = b\]
Substituting \[{5^x} = a\] and \[{3^x} = b\] again, we have
\[ \Rightarrow {5^x} = {3^x}\]
Rewriting the equation, we have
\[ \Rightarrow \dfrac{{{5^x}}}{{{3^x}}} = 1\]
By using the exponential formula, \[\left( {\dfrac{{{a^m}}}{{{b^m}}}} \right) = {\left( {\dfrac{a}{b}} \right)^m}\], we have
\[ \Rightarrow {\left( {\dfrac{5}{3}} \right)^x} = 1\]
We know that the exponent raised to the power zero is one. So, we have
\[ \Rightarrow {\left( {\dfrac{5}{3}} \right)^x} = {\left( {\dfrac{5}{3}} \right)^0}\]
We can compare the powers if the exponents are equal. On comparing the powers, we get
\[ \Rightarrow x = 0\]
Therefore, the equation \[{125^x} + {45^x} = 2{(27)^x}\] has one solution.
Hence, option B is the correct option.

Note: We can get to know the number of solutions by checking the discriminant of a quadratic equation. If the discriminant is positive \[{b^2} - 4ac > 0\], then the quadratic equation has two solutions. If the discriminant is equal \[{b^2} - 4ac = 0\], then the quadratic equation has one solution. If the discriminant is negative \[{b^2} - 4ac < 0\], then the quadratic equation has no solution. But this method is not applicable to this problem since the variable is in the power.