Hi,
Can someone please tell me how to solve the attached logarithmic problem ? the answer should be 5.
Attached is also my working.
Thanks
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Hi,
Can someone please tell me how to solve the attached logarithmic problem ? the answer should be 5.
Attached is also my working.
Thanks
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.
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 $\displaystyle x = -1$ ?
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 :PQuote:
Originally Posted by aritech
ok ... now I see it.Quote:
Originally Posted by aritech
Hi,
Can someone tell me how to work the binomial attached??
the answer should be 34749 p^8q^5
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.
From where did you bring the 13/8 ??
Attached is the full question and working:
Where are your binomial coefficients?Quote:
Originally Posted by aritech
you need to review the general binomial expansion ...
http://upload.wikimedia.org/math/d/4...1e2646d2c8.png
