Thread: 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

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 ...



not ...





why can't ?

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) ??

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

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.

Hi,

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

the answer should be 34749 p^8q^5

your binomial coefficient is incomplete ...







... 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:

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 ...