Question

How to solve this, ${4^x} + {9^x} + {25^x} = {6^x} + {10^x} + {15^x}$.

Answer 1

Hint: First prime factorizes the base of each term. After that use the law of exponent, ${\left( {{{\left( a \right)}^b}} \right)^c} = {a^{bc}}$ and ${\left( {a \times b} \right)^c} = {a^c} \times {b^c}$. After that assume the terms and substitute. Then move each term on one side. Now, multiply the whole term by 2. After that reduce the terms by using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Then do simplification to get the desired result.

Complete step-by-step solution:
We can write ${4^x}$ as,
$ \Rightarrow {4^x} = {\left( {{2^2}} \right)^x}$
We know that,
${\left( {{{\left( a \right)}^b}} \right)^c} = {a^{bc}}$
Using the rules, we get
$ \Rightarrow {4^x} = {2^{2x}}$
We can also write ${9^x}$ as,
$ \Rightarrow {9^x} = {\left( {{3^2}} \right)^x}$
Using the above rules, we get
$ \Rightarrow {9^x} = {3^{2x}}$
We can also write ${25^x}$ as,
$ \Rightarrow {25^x} = {\left( {{5^2}} \right)^x}$
Using the above rules, we get
$ \Rightarrow {25^x} = {5^{2x}}$
We can write ${6^x}$ as,
$ \Rightarrow {6^x} = {\left( {2 \times 3} \right)^x}$
We know that,
${\left( {a \times b} \right)^c} = {a^c} \times {b^c}$
Using the rules, we get
$ \Rightarrow {6^x} = {2^x} \times {3^x}$
We can write ${10^x}$ as,
$ \Rightarrow {10^x} = {\left( {2 \times 5} \right)^x}$
Using the above rules, we get
$ \Rightarrow {10^x} = {2^x} \times {5^x}$
We can write ${15^x}$ as,
$ \Rightarrow {15^x} = {\left( {3 \times 5} \right)^x}$
Using the above rules, we get
$ \Rightarrow {15^x} = {3^x} \times {5^x}$
So, the expression can be written as,
$ \Rightarrow {2^{2x}} + {3^{2x}} + {5^{2x}} = {2^x} \times {3^x} + {2^x} \times {5^x} + {3^x} \times {5^x}$
Let ${2^x} = a,{3^x} = b,{5^x} = c$.
Then the expression can be written as,
\[ \Rightarrow {a^2} + {b^2} + {c^2} = ab + ac + bc\]
Move all the terms on one side,
\[ \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0\]
Multiply both sides by 2,
\[ \Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = 0\]
Multiply 2 with each term,
\[ \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0\]
Rewrite the expressions,
\[ \Rightarrow \left( {{a^2} + {b^2} - 2ab} \right) + \left( {{b^2} + {c^2} - 2bc} \right) + \left( {{c^2} + {a^2} - 2ca} \right) = 0\]
As we know, ${\left( {p - q} \right)^2} = {p^2} + {q^2} - 2pq$. Then,
\[ \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\]
Since the sum of squares is zero then each term should be zero. So,
\[ \Rightarrow {\left( {a - b} \right)^2} = 0,{\left( {b - c} \right)^2} = 0,{\left( {c - a} \right)^2} = 0\]
Take square root all both sides,
\[ \Rightarrow \left( {a - b} \right) = 0,\left( {b - c} \right) = 0,\left( {c - a} \right) = 0\]
The above-shown condition is only possible when $a = b = c$.
Take any two-term,
$ \Rightarrow a = b$
Substitute back the values,
$ \Rightarrow {2^x} = {3^x}$
Divide both sides by ${3^x}$,
$ \Rightarrow \dfrac{{{2^x}}}{{{3^x}}} = 1$
As 1 can be written as ${\left( {\dfrac{2}{3}} \right)^0}$. So,
$ \Rightarrow {\left( {\dfrac{2}{3}} \right)^x} = {\left( {\dfrac{2}{3}} \right)^0}$
Comparing the values,
$\therefore x = 0$

Hence, the value of $x$ is 0.

Note: The conceptual knowledge about exponents and laws of exponents is required. Students should always keep in mind various laws of exponents to solve these types of questions. Mistakes can be by students while applying the law of exponent.