Question

1)solve the equation log(6x-2)-1=log(x-3)
2)evaluate without using mathematical tables,the expression 2log 5-12log 16+2 log 40
3)given that x=-2,find the values of y and z for the simultaneous equations x+y-z=-1 and x-2y+z=-7
4)solve the simultaneous equations x^2+y^2=26 where x-y=4
5)solve for x in the equation 2Sin^2x-1=Cos^2x+Sinx for 0<x<360

Answer 1

thanks

Answer 2

1) Starting with the equation log(6x-2)-1=log(x-3):
We can use the property that log(a)-log(b)=log(a/b):
log((6x-2)/(x-3))=1
Now we can use the fact that log(a)=b is equivalent to a=10^b:
(6x-2)/(x-3)=10^1
6x-2=10x-30
4x=28
x=7
Therefore, the solution to the equation is x=7.

2) Using the rules of logarithms:
2log5-12log16+2log40= log(5^2) - log(16^12) + log(40^2)
= log(25/16^12) + log(1600)
= log(25/2^48) + log(16^2 * 100)
= log(25/2^48) + log(25600)
= log(25/2^48 * 25600)
Without using tables, we cannot simplify this further.

3) Substituting x=-2 into the two equations:
x+y-z=-1 becomes -2+y-z=-1, which simplifies to y-z=1
x-2y+z=-7 becomes -2-2y+z=-7, which simplifies to -2y+z=-5
We have two equations with two variables.
Multiplying the first equation by 2 and adding it to the second equation, we get:
2y+2z-2y+z=2-5
3z=-3
z=-1
Substituting z=-1 into the equation y-z=1, we get:
y-(-1)=1
y=2
Therefore, the solution to the simultaneous equations is y=2 and z=-1.

4) Starting with the equation x-y=4, we can square both sides to get:
x^2 - 2xy + y^2 = 16
Using x^2 + y^2 = 26 from the second equation and substituting y^2 = 26 - x^2, we get:
x^2 - 2x(4+x) + 26 = 0
Simplifying and solving for x, we get:
x^2 + 8x - 6 = 0
Using the quadratic formula, we get:
x = (-8 ± sqrt(100))/2
x = -4 ± 5
Therefore, x can be either x=1 or x=-9.
Substituting each value of x into x-y=4, we get:
if x=1, then y=-3
if x=-9, then y=-13
Therefore, the solutions to the simultaneous equations are (1, -3) and (-9, -13).

5) Starting with the equation 2Sin^2x-1=Cos^2x+Sinx:
Substituting the identity Cos^2x = 1 - Sin^2x, we get:
2Sin^2x - 1 = 1 - Sin^2x + Sinx
Simplifying and moving all terms to one side:
3Sin^2x - Sinx - 2 = 0
Using the quadratic formula, we get:
Sin x = (1 ± sqrt(13))/6
We only need to consider solutions between 0 and 360 degrees, so we can use inverse sine function to get:
x = Sin^-1((1+sqrt(13))/6) ≈ 51.5 degrees or x = Sin^-1((1-sqrt(13))/6) ≈ 231.5 degrees.
Therefore, the solutions to the equation are x ≈ 51.5 degrees and x ≈ 231.5 degrees.

Answer 3

You're welcome! If you have any further questions, feel free to ask.