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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- November 11th 2012, 11:10 AMaritechLogarithmic 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 AMfkfRe: 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 AMskeeterRe: Logarithmic problem
the original equation was ...

not ...

why can't ? - November 11th 2012, 11:50 AMaritechRe: 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 PMPetrusRe: Logarithmic problem
- November 11th 2012, 12:11 PMskeeterRe: Logarithmic problem
- November 18th 2012, 07:52 AMaritechRe: 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 AMskeeterRe: Logarithmic problem
your binomial coefficient is incomplete ...

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

Attached is the full question and working: - November 18th 2012, 09:31 AMskeeterRe: Logarithmic problem
Where are your binomial coefficients?

you need to review the general binomial expansion ...

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