Question

2^x - 3^x = √(6^x - 9^x)

Answer 1

$2^x-3^x=\sqrt{6^x-9^x}\\\\D:6^x-9^x\geq0 \wedge 2^x-3^x\geq0\\D: 6^x\geq 9^x \wedge 2^x\geq3^x\\D:x\leq0 \wedge x\leq 0\qquad(\text{if }a < b \text{ then } a^x < b^x\text{ for } x > 0 \text{ and } a^x > b^x \text{ for } x < 0 )\\D:x\leq 0\\\\(2^x-3^x)^2=6^x-9^x\\(2^x-3^x)^2=3^x(2^x-3^x)\\\\\text{We divide both sides by } 2^x-3^x \text{ assuming }2^x-3^x\not=0\Leftrightarrow x\not=0.\\\\2^x-3^x=3^x\\2^x=2\cdot3^x\\\ln 2^x=\ln (2\cdot3^x )\\x\ln 2=\ln 2+\ln 3^x\\x\ln 2=\ln 2+x\ln 3$

$x\ln 2-x\ln 3=\ln 2\\x(\ln 2-\ln 3)=\ln 2\\x=\dfrac{\ln 2}{\ln 2-\ln 3}=-\dfrac{\ln 2}{\ln 3-\ln 2}=-\dfrac{\ln 2}{\ln \frac{3}{2}}\approx-1.7$

For $x=0$, our initial equation is obviously true.

Therefore $x\in\left\{-\dfrac{\ln 2}{\ln \frac{3}{2}},0\right\}$.

Answer 2

Answer to the miscellaneous equation, is x=0

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)

-> (2ˣ -3ˣ)²=(3ˣ.2ˣ - 3²ˣ)

Squaring both sides,

-> 2²ˣ+3²ˣ- 2.3ˣ.2ˣ=(3ˣ.2ˣ - 3²ˣ)

->  2²ˣ+2.3²ˣ-2.3ˣ.2ˣ-3ˣ.2ˣ=0

Factorising the terms,

-> 2ˣ(2ˣ-3ˣ) -2.3ˣ(2ˣ-3ˣ)=0

-> (2ˣ-3ˣ)(2ˣ-2.3ˣ)=0

Equating the braces to zero,

x=0 is the only solution.

Therefore, x =0 is the only solution

Learn  more about miscellaneous equation: brainly.com/question/1214333

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