log and exponential questions (1 Viewer)

[spagbowl]

ok, 1) if log (base 7) 2 = 0.36 and log (base7) 3 = 0.5 then find the value of:
a) log (base 7) 14
b) log (base 7) 3.5

2)a) use simpons rule with three function values to find the area bound by the curve y = ln x, the x-axiz and the lines x = 2 and x = 4.
b) Change the subject of y = ln x to x
c) hence find the exact area in part (a).

(i can do part a and b of Q2 but i cant do part c)

3. intergrate (2(x-cubed) - (x squared) + 5x + 3)/x and solve for x = 1 and x = 2 (sorry but imagine it set up with the integration symbol and all)

please help me

 

[Air Jordan]

1a) log(base 7) 14 = log(base7) 2 + log(base7) 7 = 0.36 + 1 = 1.36
b) log(base7) 3.5 = log(base7) 7 - log(base7) 2 = 1 - 0.36 = 0.64

 

[kurt.physics]

ok, 1) if log (base 7) 2 = 0.36 and log (base7) 3 = 0.5 then find the value of:
a) log (base 7) 14
b) log (base 7) 3.5

2)a) use simpons rule with three function values to find the area bound by the curve y = ln x, the x-axiz and the lines x = 2 and x = 4.
b) Change the subject of y = ln x to x
c) hence find the exact area in part (a).

(i can do part a and b of Q2 but i cant do part c)

3. intergrate (2(x-cubed) - (x squared) + 5x + 3)/x and solve for x = 1 and x = 2 (sorry but imagine it set up with the integration symbol and all)

please help me

a) 14 = 2 x 7

.:log₇14 = log₇2 + log₇7

= 0.36 + 1
= 1.36

b)log₇3.5 = log₇(7/2)

= log₇7 - log₇2

=1 - 0.36

= 0.64

3) all you have to do is divide each term by x, then integrate all the individual terms, the only term you would have a problem with is 3/x, and the integral, S, of (3/x) is the same as

3 x S (1\x) which is 3 log_ex

Then, depending on what is ment by "solve for x = 1 and x = 2", if you mean evaluate the integral between 1 and 2, then you would simply do it as normal

 

[kwabon]

ok, 1) if log (base 7) 2 = 0.36 and log (base7) 3 = 0.5 then find the value of:
a) log (base 7) 14
b) log (base 7) 3.5

2)a) use simpons rule with three function values to find the area bound by the curve y = ln x, the x-axiz and the lines x = 2 and x = 4.
b) Change the subject of y = ln x to x
c) hence find the exact area in part (a).

(i can do part a and b of Q2 but i cant do part c)

3. intergrate (2(x-cubed) - (x squared) + 5x + 3)/x and solve for x = 1 and x = 2 (sorry but imagine it set up with the integration symbol and all)

please help me

use 4 unit or ...
it hard to explain it to u without the diagram, but i will just give u the working out, see if u figure it from there

NOTE: DRAW THE DIAGRAM FIRST, OP

y = ln x
x = e^y
integrate that bit from ln 4 to ln 2
you should get an answer of 2 units^2

so then u consider rectangles
the big rectangle - the small rectangle
big rectangle = 4*ln 4
small rectangle = 2*ln 2
the answer = big rectanle - small rectangle - 2
= 4ln 2 - 2ln 2 - 2 , simplify that and wallah bob is you uncle

 

[spagbowl]

ahh ok everything makes so much more sense now. but still i always thought that if it is an area bound by the x-axis u dont need to make x the subject of the formula