[tex] {9}^{x} + {15}^{x} = {25}^{x} [/tex]
find the value of x
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[tex] {9}^{x} + {15}^{x} = {25}^{x} [/tex]
find the value of x
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[tex]x=\dfrac{\log2-\log(\sqrt{5}-1)}{\log5-\log3}[/tex]
[tex]\;[/tex]
[tex]\Large\textrm{\underline{\underline{Question}}}[/tex]
[tex]9^{x}+15^{x}=25^{x}[/tex]
Solve for real [tex]x[/tex].
[tex]\;[/tex]
[tex]\Large\textrm{\underline{\underline{Main Topics}}}[/tex]
We need to understand the laws of exponent, quadratic formula, and logarithm to solve this question.
[tex]\;[/tex]
[tex]\large\textrm{Laws of Exponents}[/tex]
[tex]\boxed{a^{n}\cdot b^{n}=(ab)^{n}}[/tex]
[tex]\boxed{(a^{m})^{n}=a^{mn}}[/tex]
[tex]\;[/tex]
[tex]\large\textrm{Quadraic Formula}[/tex]
[tex]\boxed{ax^{2}+bx+c=0\ (a\neq0)\iff x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}[/tex]
[tex]\;[/tex]
[tex]\large\textrm{Logarithm}[/tex]
[tex]\boxed{\log_{a^{m}}b^{n}=\dfrac{n}{m}\log_{a}b}[/tex]
[tex]\;[/tex]
[tex]\Large\textrm{\underline{\underline{Solution}}}[/tex]
Assumption:
[tex]\;[/tex]
Now
[tex]9^{x}+15^{x}=25^{x}[/tex]
[tex]\Longrightarrow(3^{2})^{x}+(3\cdot5)^{x}=(5^{2})^{x}[/tex]
[tex]\Longrightarrow 3^{2x}+3^{x}\cdot 5^{x}=5^{2x}[/tex]
[By assumption, we get the following.]
[tex]\Longrightarrow A^{2}+AB=B^{2}[/tex]
[tex]\;[/tex]
Now
[tex]A^{2}+AB=B^{2}[/tex]
[Divide both sides by [tex]B^{2}[/tex].]
[tex]\Longrightarrow \left(\dfrac{A}{B}\right)^{2}+\dfrac{A}{B}=1[/tex]
[tex]\;[/tex]
Assumption:-
[tex]\;[/tex]
Now
[tex]\left(\dfrac{A}{B}\right)^{2}+\dfrac{A}{B}=1[/tex]
[tex]\Longrightarrow C^{2}+C=1[/tex]
[tex]\Longrightarrow C^{2}+C-1=0[/tex]
[Now, this is a quadratic equation in [tex]C[/tex].]
[tex]\;[/tex]
The quadratic formula is given by
[tex]\boxed{ax^{2}+bx+c=0\ (a\neq0)\iff x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}[/tex]
[tex]\Longrightarrow C=\dfrac{-1\pm\sqrt{5}}{2}[/tex]
[tex]\;[/tex]
Remark:-
[tex]\;[/tex]
Now
[tex]C=\left(\dfrac{3}{5}\right)^{x},\ \boxed{C > 0}[/tex]
[In disregarding non-real solutions.]
[tex]\Longrightarrow \left(\dfrac{3}{5}\right)^{x}=\dfrac{-1+\sqrt{5}}{2}[/tex]
[tex]\Longrightarrow x\log\dfrac{3}{5}=\log\dfrac{-1+\sqrt{5}}{2}[/tex]
[tex]\Longrightarrow x=\dfrac{\log(\sqrt{5}-1)-\log2}{\log3-\log5}[/tex]
[tex]\Longrightarrow\boxed{x=\dfrac{\log2-\log(\sqrt{5}-1)}{\log5-\log3}}[/tex]
[This is the required answer.]