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Math Help - solve the equation: with Logarithm & one word problem

  1. #1
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    solve the equation: with Logarithm & one word problem

    the stress i had a few days has went down so much since i became a member here, i actually feel prepared for my test coming up thanks to all of you. out of the 45 problems these are the last few im having serious problems with, i wish i understood these instead of bugging you guys haha

    these last 4 problems i cant get a single answer.. problems #20 #21 and #22

    problems #24 is a word problem that i cant figure out as well, help with any problems will be much appreciated.

    i scanned the packet, here are the few problems im having trouble with.

    the test is friday and i want to know the material as best as possible because the last test i failed. i normal dont fail things, its just harder for me to understand math more than others. its very frustrating
    Last edited by murkr; April 20th 2010 at 11:22 PM.
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  2. #2
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    Quote Originally Posted by murkr View Post
    the stress i had a few days has went down so much since i became a member here, i actually feel prepared for my test coming up thanks to all of you. out of the 45 problems these are the last few im having serious problems with, i wish i understood these instead of bugging you guys haha

    these last 4 problems i cant get a single answer.. problems #20 #21 and #22

    problems #24 is a word problem that i cant figure out as well, help with any problems will be much appreciated.

    i scanned the packet, here are the few problems im having trouble with.

    the test is friday and i want to know the material as best as possible because the last test i failed. i normal dont fail things, its just harder for me to understand math more than others. its very frustrating
    To #20:

    \log(3+x)-\log(x-5)=\log(5)~\implies~\log(x+3)=\log(5)+\log(x-5) ~\implies~\log(x+3)=\log(5(x-5))

    Two logarithms are equal if their arguments are equal too:

    x+3=5(x-5)

    Solve for x

    to #21

    \frac13 \log_2(x+6)=\log_8(3x) Use the base-change-formula to get equal bases for all logarithms:

    \frac13 \log_2(x+6)=\frac{\log_2(3x)}{\log_2(8)} Since \log_2(8)=3 your equation becomes:

    \frac13 \log_2(x+6)=\frac13 \log_2(3x) Therefore

    x+6=3x

    Solve for x.

    to #22:

    \log_{21}(x+6)=1-\log_{21}(x)~\implies~\log_{21}(x+6)+\log_{21}(x)=  1

    \log_{21}(x^2+6x)=\log_{21}(21) Therefore:

    x^2+6x=21

    Solve for x. Keep in mind that a logarithm is only defined for positive numbers.
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  3. #3
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    Quote Originally Posted by earboth View Post
    To #20:

    \log(3+x)-\log(x-5)=\log(5)~\implies~\log(x+3)=\log(5)+\log(x-5) ~\implies~\log(x+3)=\log(5(x-5))

    Two logarithms are equal if their arguments are equal too:

    x+3=5(x-5)

    Solve for x

    to #21

    \frac13 \log_2(x+6)=\log_8(3x) Use the base-change-formula to get equal bases for all logarithms:

    \frac13 \log_2(x+6)=\frac{\log_2(3x)}{\log_2(8)} Since \log_2(8)=3 your equation becomes:

    \frac13 \log_2(x+6)=\frac13 \log_2(3x) Therefore

    x+6=3x

    Solve for x.

    to #22:

    \log_{21}(x+6)=1-\log_{21}(x)~\implies~\log_{21}(x+6)+\log_{21}(x)=  1

    \log_{21}(x^2+6x)=\log_{21}(21) Therefore:

    x^2+6x=21

    Solve for x. Keep in mind that a logarithm is only defined for positive numbers.
    thank you so much, that was all very helpful, i cant thank you enough! very good

    i believe you made a small mistake in problem #22 though, you posted (x+6) you must have looked at the problem above, it was (x+4) i just changed the number around and got the correct answer.

    in the end i got (x+3)(x-7) and as you said only postive numbers so the answer is 3 im assuming. thank you again!!

    can anyone else help me with problems #23 and #24?
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  4. #4
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    Quote Originally Posted by murkr View Post
    thank you so much, that was all very helpful, i cant thank you enough! very good

    i believe you made a small mistake in problem #22 though, you posted (x+6) you must have looked at the problem above, it was (x+4) i just changed the number around and got the correct answer.

    in the end i got (x+3)(x-7) <<<<< here is your mistake - so we are quit
    and as you said only postive numbers so the answer is 3 im assuming. thank you again!!

    can anyone else help me with problems #23 and #24?
    Thanks for spotting my mistake!

    You probably got:

    x^2+4x-21=0~\implies~(x-3)(x+7)=0

    which yields x = 3 \text{ or } x=-7

    but only x = 3 is a valid answer as you stated correctly
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  5. #5
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    Number 24 isn't really a word problem. They are asking you to solve the equation 1+ 1.5 ln(x+1)= 10. As always, you solve an equation by "backing out": subtract 1 from both sides to get 1.5 ln(x+1)= 9. Divide both sides by 1.5 to get ln(x+1)= 9/1.5= 6.

    Notice that each step has been an "inverse"- the opposite of some operation. Initially, we had 1 added on the left so we do the opposite- subtract 1 from both sides. Then we had 1.5 multiplying the function of x so again we do the opposite- divide both sides by 1.5. Now we have a natural logarithm. What is the "opposite" of that? The exponential! ln(x) and e^x are "inverse functions": ln(e^x)= x and e^{ln(x)}= x.

    Taking the exponential of both sides, e^{ln(x+1)}= x+ 1= e^6.

    Finally, of course, subtract 1 from both sides. x= e^6- 1. If you want a decimal (but only approximate) answer, use a calculator.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Number 24 isn't really a word problem. They are asking you to solve the equation 1+ 1.5 ln(x+1)= 10. As always, you solve an equation by "backing out": subtract 1 from both sides to get 1.5 ln(x+1)= 9. Divide both sides by 1.5 to get ln(x+1)= 9/1.5= 6.

    Notice that each step has been an "inverse"- the opposite of some operation. Initially, we had 1 added on the left so we do the opposite- subtract 1 from both sides. Then we had 1.5 multiplying the function of x so again we do the opposite- divide both sides by 1.5. Now we have a natural logarithm. What is the "opposite" of that? The exponential! ln(x) and e^x are "inverse functions": ln(e^x)= x and e^{ln(x)}= x.

    Taking the exponential of both sides, e^{ln(x+1)}= x+ 1= e^6.

    Finally, of course, subtract 1 from both sides. x= e^6- 1. If you want a decimal (but only approximate) answer, use a calculator.
    ok thank you so much! that really helped me ALOT


    for #23 i got 219 as the answer.. but i cant find what point is on the graph of G.. answers A and C both have 219.. but i cant figure out the second point. thanks again in advance, you are a really big help for me
    Last edited by murkr; April 20th 2010 at 05:52 PM.
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