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November 2nd, 2014, 03:10 PM  #1 
Newbie Joined: Oct 2014 From: Canada Posts: 7 Thanks: 0  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 midterm 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 halflife 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)=1.2511.251(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. 
November 2nd, 2014, 04:49 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,186 Thanks: 1648 
Can you post your attempts? By the way, the halflife of Co60 is nothing like 15.83 years (5.27 years would be about right). 
November 2nd, 2014, 06:15 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, Confusion9! Here is some help . . . Quote:
$\qquad\quad =\;\log\left(x^{\frac{1}{2}}\cdot x^{\frac{5}{3}}\right) + \log\left(\frac{1}{z}\right) \;=\; \log\left(x^{\frac{13}{6}}\right) + \log\left(\frac{1}{z}\right)$ $\qquad\quad=\;\log\left(x^{\frac{13}{6}}\cdot \frac{1}{z}\right) \;=\;\log\left(\frac{x^{\frac{13}{6}}}{z}\right)$ Quote:
$\quad 2(3^{3x})  3^{2x+2} \:=\:0 \quad\Rightarrow\quad 3^{2x}\left(2\cdot3^x  3^2\right) \:=\:0$ 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:
$\log_5x \:=\:\dfrac{\ln x}{\ln 5}$ $\log_{25}x \:=\:\dfrac{\ln x}{\ln25} \:=\:\dfrac{\ln x}{\ln(5^2)} \:=\:\dfrac{\ln x}{2\ln 5}$ $\log_{125}x \:=\:\dfrac{\ln x}{\ln125} \:=\:\dfrac{\ln x}{\ln(5^3)} \:=\:\dfrac{\ln x}{3\ln 5}$ The equation becomes: $\displaystyle\:\frac{\ln x}{\ln 5} + \frac{\ln x}{2\ln 5} + \frac{\ln x}{3\ln 5} \:=\:33$ $\displaystyle\qquad \frac{6\ln x}{6\ln 5} + \frac{3\ln x}{6\ln 5} + \frac{2\ln x}{6\ln 5} \:=\:33 \quad\Rightarrow\quad 6\ln x + 3\ln x + 2\ln x \:=\:33\!\cdot\!6\ln 5$ $\qquad 11\ln x \:=\:198\ln 5 \quad\Rightarrow\quad \ln x \:=\:18\ln 5$ $\qquad \ln x \:=\:\ln\left(5^{18}\right) \quad\Rightarrow\quad x \:=\:5^{18}$ Last edited by skipjack; November 2nd, 2014 at 06:56 PM.  
November 2nd, 2014, 06:48 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,186 Thanks: 1648 
(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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express as a single logarithm. simplify, if possible. ${log}_{\frac{1}{3}} 27 {log}_{\frac{1}{3}}\frac{1}{9}$,what is the simplified form of the expression \frac{4 {\frac{1}{2}}\cdot{37}}{3}?,"1.251×10" years how long
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