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Tutors Answer Your Questions about logarithm (FREE)
Question 22299: Can anyone help me with this?
I just need to sketch the graph of the following function, but need to also
know how it's done.
f(x) = 3- In x
The answer shows it as intersecting the x axis at (20,0)
Domain x>0
V. axymptote = x axis x = 0
Can anyone help?
Thanks
Sandy
Click here to see answer by stanbon(26252)   |
Question 22300: Can anyone help me solve this problem?
log (x^3 + 1) to the base of 3 - log (x+1) to the base of 3 = 1
I tried log to the base of 3 times (x^3 + 1)/(x+1) = 1
(x^3 + 1)/(x+1) = 3^1
(x^3 + 1) = 3^[1(x+1)]
and now I'm stuck, can anyone help?
Thanks
Sandy
Click here to see answer by stanbon(26252) |
Question 22453: Can anyone help with this?
[In times the cuberoot of e^2] [log 125 to the base of 5]
I did:
[In(e^2)^1/3] [log 5^3 to the base of 5]
[In e^2/3] [3log 5 to the base of 5]
[1^2/3] [3(1)]
the cube root of 1^2(3)
the answer is 2, what did I do wrong and/or how do I finish?
thanks,
Sandy
Click here to see answer by stanbon(26252) |
Question 22585: Could anybody help me prove algebraically why log based x of 3 is equal to log based 2x of 6?
Multiple answers are appreciated as our teacher suggested there is more than one way to prove. Thanks in advance!
Click here to see answer by stanbon(26252) |
Question 22747: I need help figuring out which way to proceed to solve this problem.
log2(x-2) +5 = 8- log2 4. The 2 is the base of the log. I think I should rewrite as log2(x-2) + log2 4 = 3 What is the correct way. I think answer is
4(x-2) =2^3 x=4
Thanks
Click here to see answer by Earlsdon(4898) |
Question 22799: I do apolgize. I just submitted a problem but I guess I didn't write it correctly. I'm still very new at this.
the question is
4^(x+1)=64 the x+1 are in exponential form. Where appropriate, include approximations to the nearest thousandth.
Again thank you so much.
Click here to see answer by stanbon(26252) |
Question 22799: I do apolgize. I just submitted a problem but I guess I didn't write it correctly. I'm still very new at this.
the question is
4^(x+1)=64 the x+1 are in exponential form. Where appropriate, include approximations to the nearest thousandth.
Again thank you so much.
Click here to see answer by khwang(438) |
Question 22964: HELP =) !
resolving the Question 2570 (p200 - 250) : log of 9 to the base square root of x - log of 3 to the base x = 9 + 6 log of x to the base square root of 3
a tutor found x = 3^(1/4) or x = 1/3
I only found one answer x = 3^(1/4)
Who is right ? And where is the (my?) flaw ?
Thanks a lot !!!
Here is my answer :
logV""x(9) - logx(3) = 9 + 6logV""3(x)
(logx 9) / (logx V""x) - logx 3 = 9 + 6( (logx x) / (logx V""3) )
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6( 1 / ( (logx 3) / 2 ) )
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6 / ( (logx 3) / 2 )
2logx 9 - logx 3 = 9 + 6 ( 2 / logx 3 )
2logx 9 - logx 3 = 9 + 12 / logx 3
logx 3 ( 2logx 9 - logx 3 - 9 ) = 12
logx 3 ( 4logx 3 - logx 3 - 9 ) = 12
logx 3 ( 3logx 3 - 9 ) = 12
3(logx 3)^2 - 9logx 3 = 12
(logx 3)^2 - 3logx 3 = 4
logx 3 - 3 = 4 / (logx 3)
logx 3 - 3 = 1/4 logx 3
logx 3 - 3 = logx 3^(1/4)
logx 3 - logx 3^(1/4) = 3
logx (3/3^(1/4) ) = 3
x^3 = 3/3^(1/4)
x^3 = 3^(3/4)
x^3 = 3^( (1/4) * 3 )
x^3 = ( 3^(1/4) )^3
x = 3^(1/4)
Click here to see answer by rapaljer(3610) |
Question 22964: HELP =) !
resolving the Question 2570 (p200 - 250) : log of 9 to the base square root of x - log of 3 to the base x = 9 + 6 log of x to the base square root of 3
a tutor found x = 3^(1/4) or x = 1/3
I only found one answer x = 3^(1/4)
Who is right ? And where is the (my?) flaw ?
Thanks a lot !!!
Here is my answer :
logV""x(9) - logx(3) = 9 + 6logV""3(x)
(logx 9) / (logx V""x) - logx 3 = 9 + 6( (logx x) / (logx V""3) )
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6( 1 / ( (logx 3) / 2 ) )
(logx 9) / ( (logx x) / 2 ) - logx 3 = 9 + 6 / ( (logx 3) / 2 )
2logx 9 - logx 3 = 9 + 6 ( 2 / logx 3 )
2logx 9 - logx 3 = 9 + 12 / logx 3
logx 3 ( 2logx 9 - logx 3 - 9 ) = 12
logx 3 ( 4logx 3 - logx 3 - 9 ) = 12
logx 3 ( 3logx 3 - 9 ) = 12
3(logx 3)^2 - 9logx 3 = 12
(logx 3)^2 - 3logx 3 = 4
logx 3 - 3 = 4 / (logx 3)
logx 3 - 3 = 1/4 logx 3
logx 3 - 3 = logx 3^(1/4)
logx 3 - logx 3^(1/4) = 3
logx (3/3^(1/4) ) = 3
x^3 = 3/3^(1/4)
x^3 = 3^(3/4)
x^3 = 3^( (1/4) * 3 )
x^3 = ( 3^(1/4) )^3
x = 3^(1/4)
Click here to see answer by venugopalramana(3286)  |
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