Are these solutions correct?

Solution 1

y = log base 3 (8x^7 + x)
u = 8 x^7 + x
dy/dx = 56 x^6 + 1
= (56 x^6 + 1) / (8 x^7 + x) log base 3 e

Solution 2
P = ln 6s^3 + s
u = 6s^3
dy/dx = 18 s^2
= (18s^2)/(6s^3) + 1

Solution 2
P = ln 6s^3 + s
u = 6s^3 + s
dy/dx = 18 s^2 + 1
= (18s^2 + 1)/(6s^3 + s)

Solution 3
y = 15 x^3 + 7/(2x^3) - x
y = 15x3 + 7 *2 x^-3 - x
dy/dx = 45 x^2 + 42 x^-4 - 1

Solution 3
y = 15 x^3 + 7/(2x^3) - x
dy/dx = 45 x^2 + 7/6 x^-2 - 1

Log checking

Problem 1
log (2x^2 + 6x) - 6 = log (2x)

What can I do now?


Problem 2
Am I going correct?

log (x + 6) = 2 - log (4x)
log (x + 6) = log 100 - log (4x)

Application of simple differentiation

Question 1
Water is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.

Question 2
The velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.

Example 3
The electric field E at a distance r from a point charge is E = k/r^2, where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.

Example 4
The distance s (in m) traveled by a subway train after the brakes are applied is given by s = 20t - 2t^2. How far does it travel, after the brakes are applied, in coming to a stop?

Examining the accuracy of logs

Look at the following and made some comments to assist. Remember the log base is in [] brackets

Example 1
log [8] (x + 2) = 2 - log [8] 2
log [8] 1 + log [8] 2 = 2 / (x + 2)

Example 2
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log (2x + 4) = 2
log (2x + 8) = 2
log 8
2x + 4 = 1.8
2x = 1.8 - 4

Example 3
log [8] (x + 2) = 2 - log [8] 2
log [8] x + log [8] 2 = 2 - log [8] 2
log [8] x + 2 log [8] 2 = 2

Example 4
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) = 1.67
log [8] 1.67 = x + 2

Example 5
log [8] (x + 2) = 2 - log [8] 2
log [8] (x + 2) + log [8] 2 = 2
log [8] 2(x + 2) = 2
log [8] (2x + 4) = 2
log [8] 2x = 2 - 4

Example 6
log [8] (x + 2) = 2 - log [8] 2
log (x + 2) / log 8 = 2 - log 2 / log 8
log (x + 2) / 0.903 = 2 - 0.33
log (x + 2) 0.903 * 1.667
log x + 2 = 1.5
log x = 1.5 - 2
log x = 0.5
x = 0.32

The use of ( )

What is the use of a bracket?
4 (x + 3)
4 (x + 3) (x + 3)
4 (x + 3)squared

When can you open out a bracket and when is it not allowed?

What is the differences in the following?
log 4
log 4x
log 4x + 3
log (4x + 3)
log (4x + 3)squared

Simple Algebraic Expressions

What will you say to students who do the following in an attempt to help them?


Example 1
1.14 x + 3.42 = 2x - 1
2x - 1.14x - 1 - 3.42 = 0

Example 2
1.338x - 0.699 = 0.778x + 2.334
1.338x + 0.778x = 2.334 - 0.699

Exponents

Below show some exponent approaches. This time the power will be in brackets i.e. ( )

Example 1
5 (2x - 1) = 6 (x + 3)
Multiply left by 6
Multiply right by 5
30 (2x - 1) = 30 (x + 3)
2x -1 = x + 3

Example 2
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
log 2x - 1 / log 5 = log x + 3 / log 6
12x - 6 = 5x + 15

Example 3
5 (2x - 1) = 6 (x +3)
log (5) 2x - 1 = log (6) x + 3
2x - 1 log 5 = x + 3 log 6
2x - 1 log 5 - x + 3 log 6 = 0
log [5] (2x - 1) + log [6] (-x + 3) = 0
log 10x - 5 + 9 - x = 0
log 9x + 9 = 0
log 9x = 9