# few questions

• June 28th 2008, 01:07 PM
Problem_Solver11
few questions
logarithms
(in each case check your answer by subtituting back the value for x)

$6.5^x =8.4^{(4x-10)}$

differentials

$s=4e^{3t} -e^{-2.5t}$ w.r.t. t

$y=4sin2\theta-3cos4\theta w.r.t. \theta$

integrate

$\int(2x^3-3x^2+5x-2)dx$

thanks(Itwasntme)
• June 28th 2008, 01:56 PM
Soroban
Hello, Problem_Solver11!

I see that you're still learning LaTeX . . .

Quote:

$6.5^x \:=\:8.4^{4x-10}$

Take logs: . $\ln\left(6.5^x\right) \;=\;\ln\left(8.4^{4x-10}\right) \quad\Rightarrow\quad x\ln(6.5) \;=\;(4x-10)\ln(8.4)$

. . . . $x\ln(6.5) \;=\;4x\ln(8.4) - 10\ln(8.4) \quad\Rightarrow\quad 4x\ln(8.4) - x\ln(6.5) \;=\;10\ln(8.4)$

Factor: . $x\bigg[4\ln(8.4) - \ln(6.5)\bigg] \;=\;10\ln(8.4) \quad\Rightarrow\quad x \;=\;\frac{10\ln(8.4)}{4\ln(8.4) - \ln(6.5)}$

Therefore: . $x \;=\;3.20462545\hdots$

• June 28th 2008, 02:13 PM
Problem_Solver11
yeh haha I am I only just learnt it well still are, once I keep doing them im sure I'll be ok

and thanks ^

EDIT: in differentials the s=4e3t the t is ment to be low case like the 3 how do I do this :X
Second EDIT: just done the LaTeX right now :)
• June 29th 2008, 03:12 AM
Simplicity
Quote:

Originally Posted by Problem_Solver11
logarithms
(in each case check your answer by subtituting back the value for x)

$6.5^x =8.4^{(4x-10)}$

differentials

$s=4e^{3t} -e^{-2.5t}$ w.r.t. t

$y=4sin2\theta-3cos4\theta w.r.t. \theta$

integrate

$\int(2x^3-3x^2+5x-2)dx$

thanks(Itwasntme)

$s=4e^{3t} -e^{-2.5t}$
$\frac{\mathrm{d}(e^{ax})}{\mathrm{d}x} = ae^{ax}$ Where $a$ is a constant.
$\frac{\mathrm{d}s}{\mathrm{d}t} = 12e^{3t} + 2.5e^{-2.5t}$

$y=4sin2\theta-3cos4\theta$
For Such, $\frac{\mathrm{d}(\sin Ax)}{\mathrm{d}\theta} = A\cos Ax$, $\frac{\mathrm{d}(\cos Ax)}{\mathrm{d}\theta} = - A\sin Ax$
$\frac{\mathrm{d}y}{\mathrm{d}\theta} = 8\cos 2\theta + 12 \sin 4 \theta$

$\int(2x^3-3x^2+5x-2)dx$
For Such, $\int x^a \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + c$
$= \frac{2x^4}{4} + \frac{3x^3}{3} + \frac{5x^2}{2} - \frac{2x}{1} + c$
$= \frac{x^4}{2} + 3x^3 + \frac{5x^2}{2} - 2x + c$
• June 29th 2008, 03:21 AM
Problem_Solver11
Thank you very much:)
• June 29th 2008, 03:22 AM
flyingsquirrel
Hi

Quote:

Originally Posted by Air
[..]
For Such, ${ \color{red}\int}t^a{\color{red}\,\mathrm{d}t} = \frac{x^{a+1}}{a+1}{\color{red}+C}$
[...]

• June 29th 2008, 06:53 AM
Problem_Solver11
Quote:

Originally Posted by flyingsquirrel
Hi

why did u quote her work? was there a mistake?
• June 29th 2008, 06:57 AM
flyingsquirrel
Quote:

Originally Posted by Problem_Solver11
why did u quote her work? was there a mistake?

Yes, what's in red in the quotation was missing... but you can't see it now since Air has edited her post to correct it.
• June 29th 2008, 06:58 AM
Simplicity
Quote:

Originally Posted by Problem_Solver11
why did u quote her work? was there a mistake?

When writing the general form for integral, I wrote:

$x^a = \frac{x^{a+1}}{a+1}$

and it would have made proper sense to write:

$\int x^a \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + c$

Hence I changed it.
• June 29th 2008, 07:00 AM
Problem_Solver11
Quote:

Originally Posted by Air
When writing the general form for integral, I wrote:

$x^a = \frac{x^{a+1}}{a+1}$

and it would have made proper sense to write:

$\int x^a \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + c$

Hence I changed it.

oh ok I see