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2 years ago

What is the true solution to the equation below? ln e^lnx^2=2 ln 8?

Mathematics

2 answers:

makkiz [27]2 years ago

8 0

The answer is x=4 which is B in edg

arlik [135]2 years ago

6 0

$x\ \textgreater \ 0\\\\
\ln e^{\ln x}+\ln e^{\ln x^2}=2\ln 8\\
\ln x+\ln x^2=\ln 8^2\\
\ln (x\cdot x^2)=\ln 64\\
\ln x^3=\ln 64\\
x^3=64\\
x=4\Rightarrow \text {B}
$

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Let K and T be the current ages of two siblings, Katie and Thomas. Katie is currently twice the age of Thomas. In 6 years, Katie

Pavlova-9 [17]

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Given that we should let K and T be the current ages of two siblings, Katie and Thomas.

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1 year ago

A ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is h-48t -16t^2.

Marat540 [252]

Answer:

36 feet.

Step-by-step explanation:

We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is $h(t)=-48t -16t^2$ . We are asked to find the maximum height of the ball.

We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.

To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.

Let us find x-coordinate of parabola using formula $x=-\frac{b}{2a}$ .

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$x=-\frac{48}{32}$

$x=-\frac{3}{2}$

So, the x-coordinate of the parabola is $-\frac{3}{2}$ . Now, we will substitute $x=-\frac{3}{2}$ in our given equation to find y-coordinate of parabola.

$h(t)=-48t -16t^2$

$h(-\frac{3}{2})=-48(-\frac{3}{2})-16(-\frac{3}{2})^2$

$h(-\frac{3}{2})=-24(-3)-16(\frac{9}{4})$

$h(-\frac{3}{2})=72-4*9$

$h(-\frac{3}{2})=72-36$

$h(-\frac{3}{2})=36$

Therefore, the maximum height of the ball is 36 feet.

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