# solve the equation: with Logarithm & one word problem

• Apr 19th 2010, 11:40 PM
murkr
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 :(
• Apr 20th 2010, 12:51 AM
earboth
Quote:

Originally Posted by murkr
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.
• Apr 20th 2010, 01:18 AM
murkr
Quote:

Originally Posted by earboth
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 (Clapping)

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?
• Apr 20th 2010, 08:15 AM
earboth
Quote:

Originally Posted by murkr
thank you so much, that was all very helpful, i cant thank you enough! very good (Clapping)

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 (Clapping)
• Apr 20th 2010, 10:02 AM
HallsofIvy
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.
• Apr 20th 2010, 06:35 PM
murkr
Quote:

Originally Posted by HallsofIvy
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