Algebra Pre-Algebra and Basic Algebra Math Forum

 November 2nd, 2014, 04:10 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 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 05:43 PM. November 2nd, 2014, 05:49 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,113 Thanks: 2327 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 November 2nd, 2014, 07:15 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, Confusion9!

Here is some help . . .

Quote:
 2) Express the following as a single logarithm expression: $\quad \log x^{\frac{1}{2}}+3\log z - \log z^4+5\log(\sqrt{x})$
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: $\displaystyle\:\frac{\ln x}{\ln 5} + \frac{\ln x}{2\ln 5} + \frac{\ln x}{3\ln 5} \:=\:33$

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

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

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

Last edited by skipjack; November 2nd, 2014 at 07:56 PM. November 2nd, 2014, 07:48 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,113 Thanks: 2327 (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). $\displaystyle 33*6 = 6\log_5(x) + 3\log_5(x) + 2\log_5(x) = 11\log_5(x),\text{ so }\log_5(x) = 18.$ Thanks from Confusion9 Tags confused, have an exam soon, homework help, please help ,

,

### "1.251×10" years how long

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Josh Algebra 2 November 13th, 2013 02:38 PM Shamieh Calculus 3 September 7th, 2013 05:27 PM SarahW Algebra 1 May 14th, 2013 01:50 PM Ballser Academic Guidance 60 January 14th, 2012 06:23 PM charlieboon Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      