1)Without using mathematical tables or calculators,evaluate:2log5-3log2+5log2
2)log32+log128-log729÷log33+log2-log27
3)log(3x+8)-3log2=log(x-4)
4)Find the value of x that satisfies the equation; log(x+5)=log4-log(x+2)
5)Solve for x in;log(x+24)=2log3+log(9-2x)
6)Find the value of x in the equation; log(x^2-9)=3log2+1
7)Solve for x if;(log x)^2-1/2logx=3/2
8)Without using mathematical tables or calculators,simplify;[ log(4/9)÷log(27/8)]+1
9)Solve for x in the equation; [log(4)x]^2=log(16)x^5+6
10)Evaluate; 2log5-1/2log16÷2log40
11)Solve the simultaneous equation;
log(x-1)+2log y=2log3
log x+log y=log6
12)Solve for y in the equation;log(4)y+log(y)4=2
1) 2log5 - 3log2 + 5log2
= 2log5 - (3-5)log2
= 2log5 + 2log2
= 2(log5 + log2)
= 2log(5 * 2)
= 2log10
= 2
2) log32 + log128 - log729 ÷ log33 + log2 - log27
= log(32 * 128 * 2) - log(729/27) + log2
= log(8192) - log(27) + log2
= log(8192/27 * 2)
= log(512 * 2)
= log(1024)
= 3
3) log(3x+8) - 3log2 = log(x-4)
log((3x+8)/8) = log(x-4)
(3x+8)/8 = x-4
3x + 8 = 8x - 32
32 - 8 = 8x - 3x
24 = 5x
x = 24/5
4) log(x+5) = log4 - log(x+2)
log(x+5) = log(4/(x+2))
x+5 = 4/(x+2)
x^2+7x+10 = 4
x^2+7x+6 = 0
(x+1)(x+6) = 0
x = -1 or x = -6
5) log(x+24) = 2log3 + log(9-2x)
log(x+24) = log(3^2) + log(9-2x)
log(x+24) = log(9(9-2x))
x+24 = 9(9-2x)
x+24 = 81-18x
19x = 57
x = 3
6) log(x^2-9) = 3log2 + 1
log(x^2-9) = log(8) + log(10)
log(x^2-9) = log(8 * 10)
log(x^2-9) = log(80)
x^2-9 = 80
x^2 = 89
x = sqrt(89)
7) (log x)^2 - 1/2logx = 3/2
(log x)^2 - (1/2)logx - 3/2 = 0
Let u = log x
u^2 - (1/2)u - 3/2 = 0
2u^2 - u - 3 = 0
(2u+3)(u-1) = 0
u = -3/2 or u = 1
log x = -3/2 or log x = 1
x = 10^(-3/2) or x = 10^1
x = 0.0316 or x = 10
8) [ log(4/9) ÷ log(27/8) ] + 1
log(4/9) = log2^2 - log3^2
= 2log2 - 2log3
log(27/8) = log3^3 - log2^3
= 3log3 - 3log2
[ 2log2 - 2log3 ÷ 3log3 - 3log2 ] + 1
= [ 2log2 - 2log3 ]+1
= 2log2 - 2log3 + 1
9) [log(4)x]^2 = log(16)x^5 + 6
(log(4)x)^2 = log(2^4)x^5 + log(10^6)
= 4log2x^5 + 6
(log(4)x)^2 - 4log2x^5 = 6
Let u = log(4)x
u^2 - 4u + 6 = 0
This is a quadratic equation that doesn't have real solutions, so there are no solutions for x.
10) 2log5 - 1/2log16 ÷ 2log40
= 2log5 - 1/2(log2^4) ÷ 2log(2^3 * 5)
= 2log5 - 1/2(4log2) ÷ 2(3log2 + log5)
= 2log5 - 2log2 ÷ 6log2 + 2log5
= 2(log5 - log2 ÷ 3log2 + log5)
= 2log(5/2) ÷ 3log(2/5)
= log((5/2)^2) ÷ log((2/5)^3)
= log(25/4) ÷ log(8/125)
= log(25/4) ÷ log(1/(125/8))
= log(25/4) ÷ log(8/125)
= log(25/4) ÷ log(2^3/5^3)
= log(25/4) ÷ 3(log2 - log5)
= log(25/4) ÷ 3(log(2/5))
= log(25/4) ÷ log(2/5) - log5)
= log(25/4) ÷ log((2/5) * (1/5))
= log(25/4) ÷ log(2/25)
= log((25/4) * (25/2)) ÷ log2
= log(625/8) ÷ log2
= log2(625/8) ÷ log2
= log2(625/8)
= log2(625) - log2(8)
= 9 - 3
= 6
11) log(x-1) + 2log y = 2log3
log(x-1) + log(y^2) = log3^2
log((x-1)y^2) = log9
(x-1)y^2 = 9
xy^2 - y^2 = 9
y^2(x - 1) = 9
log x + log y = log6
log(xy) = log6
xy = 6
Using the second equation, we can express x in terms of y: x = 6/y
Substituting this into the first equation:
(6/y - 1)y^2 = 9
6y - y^2 = 9
-y^2 + 6y - 9 = 0
y^2 - 6y + 9 = 0
(y-3)^2 = 0
y = 3
Substituting y = 3 into x = 6/y:
x = 6/3
x = 2
Therefore, the simultaneous equations are satisfied when x = 2 and y = 3.
12) log(4)y + log(y)4 = 2
log4(y^2) + log(y)4 = 2
log4(y^2 * 4^4) = 2
log4(y^2 * 256) = 2
log4(256y^2) = 2
256y^2 = 4^2
256y^2 = 16
y^2 = 16/256
y^2 = 1/16
y = √(1/16)
y = 1/4
Therefore, y = 1/4.
1) To evaluate 2log5 - 3log2 + 5log2:
You can use logarithmic properties to simplify the expression. The property log(a) - log(b) = log(a/b) allows us to combine the log2 terms. Similarly, the property clog(a) = log(a^c) allows us to rewrite 5log2 as log(2^5).
So, the expression becomes:
2log5 - 3log2 + 5log2 = log(5^2) - log(2^3) + log(2^5)
= log(25) - log(8) + log(32)
At this point, we can use the property log(a) + log(b) = log(ab) to combine the terms inside the logarithm.
= log(25 * 32 / 8)
= log(800 / 8)
= log(100)
Now, since 10^2 = 100, we can conclude that log(100) = 2.
Therefore, the value of the given expression is 2.
2) To evaluate log32 + log128 - log729 ÷ log33 + log2 - log27:
Again, you can use logarithmic properties to simplify the expression. The property log(a) - log(b) = log(a/b) can be used to combine the terms involving division.
So, the expression becomes:
log32 + log128 - log729 ÷ log33 + log2 - log27 = log(32) + log(128) - log(729/33) + log(2) - log(27)
= log(32 * 128 / (729/33) * 2) - log(27)
= log(32 * 128 * 33 / 729 * 2) - log(27)
= log(165888 / 1458) - log(27)
= log(114) - log(27)
Now, using the property log(a) - log(b) = log(a / b), we can simplify further:
= log(114 / 27)
At this point, we can use the property log(a/b) = log(a) - log(b).
= log(2^1 * 3^2 / 3^3)
= log(2/3)
Therefore, the value of the given expression is log(2/3).
3) To solve the equation log(3x+8) - 3log2 = log(x-4) for x:
We can use logarithmic properties to solve for x. Start by applying the property log(a) - log(b) = log(a/b):
log(3x+8) - 3log2 = log(x-4)
log(3x+8) - log(2^3) = log(x-4)
log(3x+8) - log(8) = log(x-4)
Next, combine the logarithms with the same base using the property log(a) - log(b) = log(a/b):
log((3x+8)/8) = log(x-4)
Remove the logarithm by exponentiating both sides:
(3x+8)/8 = x-4
Now, simplify and solve for x:
3x+8 = 8(x-4)
3x+8 = 8x-32
8+32 = 8x-3x
40 = 5x
x = 40/5
x = 8
Therefore, the value of x that satisfies the equation is 8.
(Please let me know if you would like me to continue solving the remaining questions.)