# Logarithmic problem

• Nov 11th 2012, 11:10 AM
aritech
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
• Nov 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.
• Nov 11th 2012, 11:38 AM
skeeter
Re: Logarithmic problem
the original equation was ...

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

not ...

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

why can't $\displaystyle x = -1$ ?
• Nov 11th 2012, 11:50 AM
aritech
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) ??
• Nov 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
• Nov 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.
• Nov 18th 2012, 07:52 AM
aritech
Re: Logarithmic problem
Hi,

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

the answer should be 34749 p^8q^5
• Nov 18th 2012, 08:19 AM
skeeter
Re: Logarithmic problem
your binomial coefficient is incomplete ...

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

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

$\displaystyle 34749p^8q^5$

... next time, start a new problem with a new post. Do not piggy-back onto an older one.
• Nov 18th 2012, 09:11 AM
aritech
Re: Logarithmic problem
From where did you bring the 13/8 ??

Attached is the full question and working:
• Nov 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 ...