Question

NEED HELP:

Solve this equation: 3(2^x)=(0.6)^-x

Answer 1

I am waiting for you robtobey,hihi

Answer 2

Darn. I was typing it up but I accidentally deleted it. This editor isn't very useful.

Answer 3

\[\text{Log}\left[3 \left(2^x\right)\right]==\text{Log}\left[\left(\frac{6}{10}\right)^{-x}\right] \]\[x \text{Log}[2]+\text{Log}[3]= x (-\text{Log}[3]+\text{Log}[5])\]Solve for x.\[x\text{ =}-\frac{\text{Log}[3]}{\text{Log}[2]+\text{Log}[3]-\text{Log}[5]}=-6.02569 \]

Answer 4

\[3(2^{x}) = (\frac{ 3 }{ 5 })^{-x}\]
Take the logarithm of both sides.
\[\log3(2^{x}) = \log(\frac{ 3 }{ 5 })^{-x}\]
Use the laws of logarithms to expand the equation. I will split this into two steps.
\[\log3 + \log2^{x} = \log(\frac{ 3 }{ 5 })^{-x}\]
\[\log3 + x \log2 = -x \log(\frac{3}{5})\]
Now move all terms with x in them to the left, and all terms without x in them to the right.
\[x \log2 +x \log(\frac{ 3 }{ 5 }) = - \log3\]
Now solve for x...
\[x (\log2 + \log(\frac{ 3 }{ 5 })) = - \log3\]
\[x = \frac{ - \log3 }{ \log2 + \log(\frac{ 3 }{ 5 }) }\]

Plug it into your calculator and enjoy. Feel free to ask if you have any problems with the laws of logarithms. Also, you can further split the log(3/5) into log3 - log5.

Answer 5

oh ok...I see...thanks you so much guys for your help

Answer 6

I have another problem:

write the partial fraction decomposition of the given rational expression:

5x^2/(x^3-3x^2-9x-5)

Answer 7

@Interrif ,@Loser66 ,@robtobey : please kindly help me solve this problem as well...thanks

Answer 8

First you should factor the bottom. Using synthetic division (which I can explain if you need me to) you get

\[\frac{5x^{2}}{x^{3} - 3x^{2} -9x -5} = \frac{5x^{2}}{(x-4)(x-1)^{2}}\]

With Partial Fraction Decomposition, you get...

\[\frac{5x^{2}}{x^{3} - 3x^{2} -9x -5} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x-4}\]

When the denominator is the same, we can focus on the numerator, so multiply through by the various terms in the denominator to end up with...

\[5x^2 = A(x-1)(x-4) + B(x-4) + C(x-1)^2\]

Since this is valid for all x, we can choose any values of x that make it convenient to remove terms from the numerator we don't want. For x = 1, we get...

\[5(1)^2 = A(1-1)(1-4) + B(1-4) + C(1-1)^2\]
\[5 = -3B\]
\[B = \frac{ -5 }{ 3 }\]

With x = 4, you end up with...

\[5(4)^2 = A(4-1)(4-4) + B(4-4) + C(4-1)^2\]
\[80 = 9C\]
\[C = \frac{ 80 }{ 9 }\]

Solving for A is pretty annoying. But if you isolate for A and choose x = -1, you get...

\[A = \frac{ 5x^x - B(x-4) - C(x-1)^2 }{(x-1)(x-4)}\]
\[A = \frac{ 5(-1)^2 - (\frac{ -5 }{ 3 })((-1)-4) - (\frac{ 80 }{ 9 })((-1)-1)^2 }{((-1)-1)((-1)-4)}\]
\[A = \frac{ 13 }{ 9 }\]

So the final answers should be...
\[\frac{5x^{2}}{x^{3} - 3x^{2} -9x -5} = \frac{13}{9(x-1)} - \frac{5}{3(x-1)^2} + \frac{80}{9(x-4)}\]

This is only the second ever partial fraction decomposition question I have ever done, so you might want to get someone else better at math than me to check if it's right.

Answer 9

Sorry for the wait. The editor is pretty tough for me to use.

Answer 10

\[\frac{5 x^2}{x^3-3 x^2-9 x-5}=\frac{55}{36 (x+1)}-\frac{5}{6 (x+1)^2}+\frac{125}{36 (x-5)} \]\[\frac{31 x^2-80 x+64}{3 x^3-18 x^2+27 x-12}=\frac{13}{9 (x-1)}-\frac{5}{3 (x-1)^2}+\frac{80}{9 (x-4)} \]

Answer 11

Yep, I messed up at the very first step when doing synthetic division. Sorry.