Simplify to the extent possible (logx16)(log2x)

Mathematics

1 answer:

sasho [114]2 years ago

7 0

Answer:

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

Change base x to base 2:

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

You might be interested in

Stretch your thinking, solve the story problem. Draw comparison bars. There were 11 hikers here yesterday. There are 4 hikers he

Kryger [21]

11-4=7 your welcome

6 0

2 years ago

What is 5+5 jfkkckdndnd

mario62 [17]

5+5 is ten............

5 0

3 years ago

Read 2 more answers

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$