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November 2nd, 2014, 03:10 PM   #1
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Need Help Please.. Really Confused!!

I am having trouble completing these following homework questions and I would like to learn how to do them soon as I have a mid-term exam coming up in 2 days. Any help will be really appreciated. Thanks

1) Express in exponential or logarithmic form, whichever isn't given.

1a) y=27(1/3)^2x
1b) -3log_5(z)-x^2=2x+1


2) Express the following as a single logarithm expression in the most simplified form:
log(x)^1/2+3log(z)-[log(z)^4+5log(x square rooted by 3)


5) Solve for x.

5c) 2(27)^x=9^(x+1)


6) The half-life of Co60 is 15.83 years. Determine how long it takes for 3.5mg of Co60 to decay to 1mg. Round your answer to two decimals places.


7) In a particular circuit, the current, I, in amperes, can be found using the formula I(t)=1.251-1.251(10^-0.02668t), after t seconds.
7a) What is the current in the circuit after 1 minute?
7b) Rewrite the equation for I(t), solving for t.
7c) Use the equation in part b to determine the time it takes for the current to reach 0.5A.


10) Solve log_5x+log_25x+log_125x=33.


Any help would be really appreciated. If anyone who helps could please explain the steps of what they did to get the answer, that would be really useful for me to learn how to do the questions. Please just don't give the final answer as my main goal is learn the process of how to do the questions. Thanks again.

Last edited by skipjack; November 2nd, 2014 at 04:43 PM.
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November 2nd, 2014, 04:49 PM   #2
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Can you post your attempts?

By the way, the half-life of Co-60 is nothing like 15.83 years (5.27 years would be about right).
Thanks from Confusion9
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November 2nd, 2014, 06:15 PM   #3
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Hello, Confusion9!

Here is some help . . .


Quote:
2) Express the following as a single logarithm expression:

$\quad \log x^{\frac{1}{2}}+3\log z - \log z^4+5\log(\sqrt[3]{x})$
We have: $\:\log x^{\frac{1}{2}} + 5\log x^{\frac{1}{3}} + \log z^3 - \log z^4 \;=\;\log x^{\frac{1}{2}} + \log x^{\frac{5}{3}} + \log\left(\frac{z^3}{z^4}\right)$

$\qquad\quad =\;\log\left(x^{\frac{1}{2}}\cdot x^{\frac{5}{3}}\right) + \log\left(\frac{1}{z}\right) \;=\; \log\left(x^{\frac{13}{6}}\right) + \log\left(\frac{1}{z}\right)$

$\qquad\quad=\;\log\left(x^{\frac{13}{6}}\cdot \frac{1}{z}\right) \;=\;\log\left(\frac{x^{\frac{13}{6}}}{z}\right)$



Quote:
5) Solve for $x:\;(c)\;2(27)^x \:=\: 9^{x+1}$
We have: $\:2(3^3)^x \:=\:(3^2)^{x+1} \quad\Rightarrow\quad 2(3^{3x}) \:=\:3^{2x+2}$

$\quad 2(3^{3x}) - 3^{2x+2} \:=\:0 \quad\Rightarrow\quad 3^{2x}\left(2\cdot3^x - 3^2\right) \:=\:0$

Hence: $\:3^{2x} \:=\:0$ . . . not possible

$\quad$ and: $\:2\cdot 3^x- 9 \:=\:0 \quad\Rightarrow\quad 3^x \,=\,\frac{9}{2}$

Take logs: $\:\ln(3^x) \:=\:\ln\left(\frac{9}{2}\right) \quad\Rightarrow\quad x\ln 3 \:=\:\ln\left(\frac{9}{2}\right) $

Therefore: $\:x \:=\:\dfrac{\ln(\frac{9}{2})}{\ln3}$



Quote:
10) Solve:$\:\log_5x + \log_{25}x+\log_{125}x \:=\: 33$
I used the base-change formula: $\:\log_ba \:=\:\dfrac{\ln a}{\ln b}$

$\log_5x \:=\:\dfrac{\ln x}{\ln 5}$

$\log_{25}x \:=\:\dfrac{\ln x}{\ln25} \:=\:\dfrac{\ln x}{\ln(5^2)} \:=\:\dfrac{\ln x}{2\ln 5}$

$\log_{125}x \:=\:\dfrac{\ln x}{\ln125} \:=\:\dfrac{\ln x}{\ln(5^3)} \:=\:\dfrac{\ln x}{3\ln 5}$

The equation becomes: $\displaystyle\:\frac{\ln x}{\ln 5} + \frac{\ln x}{2\ln 5} + \frac{\ln x}{3\ln 5} \:=\:33$

$\displaystyle\qquad \frac{6\ln x}{6\ln 5} + \frac{3\ln x}{6\ln 5} + \frac{2\ln x}{6\ln 5} \:=\:33 \quad\Rightarrow\quad 6\ln x + 3\ln x + 2\ln x \:=\:33\!\cdot\!6\ln 5$

$\qquad 11\ln x \:=\:198\ln 5 \quad\Rightarrow\quad \ln x \:=\:18\ln 5$

$\qquad \ln x \:=\:\ln\left(5^{18}\right) \quad\Rightarrow\quad x \:=\:5^{18}$

Thanks from Confusion9

Last edited by skipjack; November 2nd, 2014 at 06:56 PM.
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November 2nd, 2014, 06:48 PM   #4
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(5c). $\displaystyle 2(27)^x=9^{x+1} \implies \log_3(2) + \log_3(27)x = \log_3(9)(x + 1) \\
\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \implies \log_3(2) + 3x = 2(x + 1) = 2x + 2$

(10). $\displaystyle 33*6 = 6\log_5(x) + 3\log_5(x) + 2\log_5(x) = 11\log_5(x),\text{ so }\log_5(x) = 18.$
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