An important rule of logs is a*log b = log b^a.
Thus, 2 (log to the base 5 of )(5x^3) = (log to the base 5 of ) (5x^3)^2, or
(log to the base 5 of ) (25x^6).
Next, (1/3) (log to the base 5 of ) (x^2+6) = (log to the base 5 of ) (x^2+6)^(1/3).
Here, the addition in the middle of the given expression indicates multiplication:
2Log5(5x^3)+1/3log5(x^2+6) = (log to the base 5 of ) { (5x^3)^2 * (x^2+6)^(1/3) }.
Here we've expressed the given log quantity as a single log.
Answer:
Step-by-step explanation:
Given the limit of a function expressed as 
, to evaluate the following steps must be carried out.
Step 1: substitute x = 0 into the function
Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20sin%28x%29-tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%28x%5E3%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7Bcos%28x%29-sec%5E2%28x%29%7D%7B3x%5E2%7D%5C%5C)
Step 3: substitute x = 0 into the resulting function
Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20cos%28x%29-sec%5E2%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%283x%5E2%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B-sin%28x%29-2sec%5E2%28x%29tan%28x%29%7D%7B6x%7D%5C%5C)
Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20-sin%28x%29-2sec%5E2%28x%29tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%286x%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E2%28x%29sec%5E2%28x%29%2B2sec%5E2%28x%29tan%28x%29tan%28x%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E4%28x%29%2B2sec%5E2%28x%29tan%5E2%28x%29%5D%7D%7B6%7D%5C%5C)
Step 7: substitute x = 0 into the resulting function in step 6
![= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}](https://tex.z-dn.net/?f=%3D%20%20%5Cdfrac%7B%5B%20-cos%280%29-2%28sec%5E4%280%29%2B2sec%5E2%280%29tan%5E2%280%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B-1-2%280%29%7D%7B6%7D%20%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B6%7D)
<em>Hence the limit of the function </em>
.
So there is 65 yards and one yard is 3 feet so you multiple 65 by 3 which is 195 . Then you divide by 4 as you have to see how many four foot pieces can be cut which you end up with 48.75 which means that you can have 48 four foot pieces. So since you have 48 four foot pieces you multiply 48 by 4 and you have 192 and if you subtract 195 which is how many feet there is in the 65 yards you get three feet and that is how much of the rope that is left over.
Well I think I can help you out ...
if you assume input x=2
the output will be as following :
y= -1/3(2)+2
=-2/3+2
=-2/3+6/3
=(-2+6)/3
=4/3
=1.333
