How to solve : 〖1/(3^(x-2).9^(3x) )=1/(〖√(〖27〗^(2x+3) ))〗^(-1) )〗^

1 answer:

6 0

$\dfrac{1}{3^{x-2}\cdot9^{3x}}=\dfrac{1}{(\sqrt{27^{2x+3}})^{-1}}\\
\dfrac{1}{3^{x-2}\cdot(3^2)^{3x}}=\dfrac{1}{(((3^3)^{2x+3})^{\frac{1}{2}})^{-1}}\\
\dfrac{1}{3^{x-2}\cdot3^{6x}}=\dfrac{1}{((3^{6x+9})^{\frac{1}{2}})^{-1}}\\
\dfrac{1}{3^{x-2+6x}}=\dfrac{1}{(3^{3x+\frac{9}{2}})^{-1}}\\
\dfrac{1}{3^{7x-2}}=\dfrac{1}{3^{-3x-\frac{9}{2}}}\\
3^{7x-2}=3^{-3x-\frac{9}{2}}\\
7x-2=-3x-\dfrac{9}{2}\\
14x-4=-6x-9\\
20x=-5\\
x=-4
$

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30w^3 -6w<br> ---------------<br> 6w

7nadin3 [17]

$\frac{30w^3-6w}{6w} = \frac{6w(5w^2-1)}{6w} =5w^2-1$

7 0

3 years ago

Read 2 more answers

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$