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3 years ago

Simplify to the extent possible (logx16)(log2x)

Mathematics

1 answer:

sasho [114]3 years ago

7 0

Answer:

${ \tt{ = ( log_{x}16)( log_{2}x) }}$

Change base x to base 2:

${ \tt{ = (\frac{ log_{2}16}{ log_{2}x } )( log_{2}x)}} \\ \\ { \tt{ = log_{2}(16) }} \\ = { \tt{ log_{2}(2) }} {}^{4} \\ = { \tt{4 log_{2}(2) }} \\ = { \tt{4}}$

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A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio

Klio2033 [76]

Answer:

k = 0.032
P0 = 131.836

Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

$P(t)=P_0\cdot e^{kt}$

We know from the given points that we can write the equation as ...

$P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t$

Comparing this to the desired form, we see that ...

$P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693$

So, the approximate equation for P is ...

$P(t)=131.836\cdote^{0.032t}$

And the parameters of interest are ...

k = 0.032
P0 = 131.836

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3 years ago

Help thanks 7.04 love u guys

Ugo [173]

Answer:

D. $-\frac{3}{4}$

Step-by-step explanation:

The given equation is

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When we divide through by 4 we get;

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Comparing to

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The corrrect choice is D.

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4 years ago

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Answer:

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