[tex]2^x-3^x=\sqrt{6^x-9^x}\\\\D:6^x-9^x\geq0 \wedge 2^x-3^x\geq0\\D: 6^x\geq 9^x \wedge 2^x\geq3^x\\D:x\leq0 \wedge x\leq 0\qquad(\text{if }a < b \text{ then } a^x < b^x\text{ for } x > 0 \text{ and } a^x > b^x \text{ for } x < 0 )\\D:x\leq 0\\\\(2^x-3^x)^2=6^x-9^x\\(2^x-3^x)^2=3^x(2^x-3^x)\\\\\text{We divide both sides by } 2^x-3^x \text{ assuming }2^x-3^x\not=0\Leftrightarrow x\not=0.\\\\2^x-3^x=3^x\\2^x=2\cdot3^x\\\ln 2^x=\ln (2\cdot3^x )\\x\ln 2=\ln 2+\ln 3^x\\x\ln 2=\ln 2+x\ln 3[/tex]
[tex]x\ln 2-x\ln 3=\ln 2\\x(\ln 2-\ln 3)=\ln 2\\x=\dfrac{\ln 2}{\ln 2-\ln 3}=-\dfrac{\ln 2}{\ln 3-\ln 2}=-\dfrac{\ln 2}{\ln \frac{3}{2}}\approx-1.7[/tex]
For [tex]x=0[/tex], our initial equation is obviously true.
Therefore [tex]x\in\left\{-\dfrac{\ln 2}{\ln \frac{3}{2}},0\right\}[/tex].
Answer to the miscellaneous equation, is x=0 and x=-1.71
Miscellaneous equation are the equations which are not polynomial
The question can be solved by:
Factorising Splitting the terms to find the required solution
Completing the squares, etc
2ˣ - 3ˣ = √(6ˣ - 9ˣ)
Squaring both sides,
2²ˣ+3²ˣ -2.2ˣ.3ˣ=2ˣ.3ˣ - 3²ˣ
2²ˣ+2.3²ˣ -3.2ˣ.3ˣ=0
Factorising the terms,
2ˣ(2ˣ-3ˣ) -2.3ˣ(2ˣ-3ˣ)=0
(2ˣ-3ˣ)(2ˣ-2.3ˣ)=0
Equating the braces to zero,
Either,
2ˣ=3ˣ
x=0
or
2ˣ=2.3ˣ
Taking log on both sides
xln2=ln2+xln3
x=ln2/(ln2-ln3)
=-1.71
Therefore, x=0 and x=-1.71 is the only solution
Learn more about miscellaneous equation: https://brainly.com/question/121433
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