Logarithmic problem

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November 11th 2012, 11:10 AM
aritech

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Logarithmic problem

Hi,

Can someone please tell me how to solve the attached logarithmic problem ? the answer should be 5.
Attached is also my working.

Thanks
November 11th 2012, 11:27 AM
fkf

Re: Logarithmic problem

log(x^2-5)-log(x) = log(4)

Use the logarithmic law that states that log(a)-log(b) = log(a/b)

log((x^2-5)/x) = log(4) (Here's were your error is in your calculation)
(x^2-5)/x = 4
x^2-5 = 4x

Solving this equation gives the solution x = -1 and x = 5.

But since negative solutions are not defined in the equation log(x) the only solution is x = 5.
November 11th 2012, 11:38 AM
skeeter

Re: Logarithmic problem

the original equation was ...

$\log\left(\frac{x^2-5}{x}\right) = \log{4}$

not ...

$\log(x^2-5) - \log{x} = \log{4}$

why can't $x = -1$ ?
November 11th 2012, 11:50 AM
aritech

1 Attachment(s)

Re: Logarithmic problem

Thanks for the reply. At skeeter, no the first one was the is the original equation. There is another attachment.

Another problem. The result of the attached problem is 6.66 while on the book it states that the result should be 6.058. What am I doing wrong? Isn't it gust (Ln14.91/(Ln1.5) ??
November 11th 2012, 12:09 PM
Petrus

Re: Logarithmic problem

Quote:

Originally Posted by aritech

Thanks for the reply. At skeeter, no the first one was the is the original equation. There is another attachment.

Another problem. The result of the attached problem is 6.66 while on the book it states that the result should be 6.058. What am I doing wrong? Isn't it gust (Ln14.91/(Ln1.5) ??

i would just take ^1/1.5 on both side on that problem :P
November 11th 2012, 12:11 PM
skeeter

Re: Logarithmic problem

Quote:

Originally Posted by aritech

Thanks for the reply. At skeeter, no the first one was the is the original equation. There is another attachment.

ok ... now I see it.
November 18th 2012, 07:52 AM
aritech

1 Attachment(s)

Re: Logarithmic problem

Hi,

Can someone tell me how to work the binomial attached??

the answer should be 34749 p^8q^5
November 18th 2012, 08:19 AM
skeeter

Re: Logarithmic problem

your binomial coefficient is incomplete ...

$\binom{13}{8} \cdot (3p)^8 \left(\frac{q}{3}\right)^5$

$\frac{13!}{8!(13-8)!} \cdot (3^3 p^8 q^5)$

$34749p^8q^5$

... next time, start a new problem with a new post. Do not piggy-back onto an older one.
November 18th 2012, 09:11 AM
aritech

1 Attachment(s)

Re: Logarithmic problem

From where did you bring the 13/8 ??

Attached is the full question and working:
November 18th 2012, 09:31 AM
skeeter

Re: Logarithmic problem

Quote:

Originally Posted by aritech

From where did you bring the 13/8 ??

Attached is the full question and working:

Where are your binomial coefficients?

you need to review the general binomial expansion ...

http://upload.wikimedia.org/math/d/4...1e2646d2c8.png