Questions
- Use mathematical table to evaluate.
- Given that y = Bxn. Make n the subject of the formula and simplify your answer
- Without using mathematical tables or calculators evaluate: 6log264 + 10log3(243)
- Find the value of x that satisfies the equation log (2x – 11) – log 2 =log 3 – log x
- Use logarithms to evaluate to 3 significant figures
- Use logarithm tables in all your steps to evaluate:
leaving your answer to four decimal places - Make L the subject in :
- Using logarithm tables solve.
- Solve the simultaneous equation:-
Log (x-1) + 2log y = 2log3
log x + log y = log 6 - Without using logarithms tables or calculator evaluate:-
4/5log1032 + log1050-3log10 - Use logarithms to evaluate:-
and express the answer in standard form - Solve for x given that :- log (3x + 8) – 3log2 = log (x-4)
- In this question, show all the steps in your calculations, giving your answer at each stage.
Use logarithms correct to 4 decimal places to evaluate: - Use logarithms to evaluate correct to 4 s.f
- Without using logarithm tables evaluate:
- Without using a calculator/mathematical tables, solve: Log8 (x + 5) – log8(x -3) = Log84
- Use tables to calculate;(6.572 + 6.57) ÷ (7.922 x 30.08)(Give your answer to 4 decimal places)
- If log2 = 0.30103, and log3 = 0.47712, calculate without using tables or calculators the value of log120
- Solve for x in the following equation; Log2(3x -4) = 1/3log28x6 – log24
- By showing all the steps, use logarithms to evaluate:
- Solve the logarithimic equation: log10(6x – 2) – 1 = log10 (x - 3)
- In this question, show all the steps in your calculations, giving your answers at each stage. Use logarithms, correct to 4 d.p to evaluate:-
- Evaluate using logarithms
Answers
-
- Log y = log B + n log x
n log x = log y – log B
n = Log (y/B)/Log x - = 6 log24 + 10 log33
= 12 log22 + 10 log33
= 12 + 10 - Log (2x - 11)/2 = log 3/x
(2x – 11) = 3/x
2x2 _ 11x -6 = 0
(2x + 1 ) (x – 6) = 0
x = - ½ or 6
x = 6 -
-
- H3 = 3d(L - d)/10L
3dL - 10H3L= 3d2
L(3d -10H3)3d2
L = 3d2/3d - 10H3 -
- Log y2 (x-1) = log 9 y2 (x-1) = 9 ….(1)
log (xy) log 6 xy = 6 ....2
from (2) x = 6/y
substitute in (1) y(6 -1)/y = 9
6y – y2 = 9
y2– 6y + 9 = 0
(y-3)2 = 0
y = 3
∴x = 2 - 4/5 log1025 + log1025x2 – log 10
4log 2 = log1025x2 – 3log2
2log10 + 2log5
Log 10 x 100 -
- Log 3x + 8 – log 8 = log (x-4)
Log (3x + 8)/8 = log (x-4)
3x + 8 = x -4
3x + 8 = 8x – 32
5x = 40 -
-
- From square roots 12.25 = 3.5
3.264 x 1.215 x 3.5x√107
1.088 x 0.4725 x 107
3264 x 1215 x35
1088 x 4725
√27 = 3 - Log8(x + 5) – log8(x -3) = Log84
Log8(x + 5)/x – 3 = log84
x + 5 = 4
x – 3
4x – 12 = x + 5
3x = 17
x = 17/3 = 52/3
Or log8x+5/x-3 = 2/3
8 2/3 = x + 5/x – 3
23(2/3) = x + 5/x -3
22 = x +5/x - 3 →4 = x + 5/x - 3
4x -12 = x + 5 3x = 17
x = 17/3 = 52/3 -
- Log 120 = log 4 + log 3 + log 10
= log22 + log3 + log 10
= 2log2 + log3 + log 10
= 2(0.30103) + 0.47712 + 1
= 2.07918 - Log2 (3x – 4) = 1/3 lo28x6 – log24
Log2 (3x – 4) = log2(23x6) - log24
Log2 (3x – 4) = log22x2 – log2 4
Log2 (3x – 4) – log2 2x2/4
= 3x – 4 = 2x2/4
2x2 – 12x + 16 = 0
x2– 6x + 8 = 0
x – 2x – 4x + 8 = 0
(x – 2) (x- 4) = 0
x = 2 or x = 4 -
- Det 2 - -3 = 5
Area of AIBICI = 5 x 15
= 75 cm2 - Log10(6x-2) – log10 = log10(x-3)
Log (6x -2)/10 = log (x-3)
6x - 2/10 = x -3
6x – 2 = 10x -30
x = 7 - No. Log
0.075262 2.8766 x 2 = 3.7532
6.652 0.8230 = 0.8230
4.9302/3 = 6 + 2.9302/3
= 2.9767
Antilog = 9.4776 x 10-2
= 0.094776(accept 0.09478)
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