1) Use the properties of logarithms to write the following expression as a single logarithm.
7 logx+1/8 logy-3/8 logz-log a
2) Find the following (use four decimal places)
3) The volume of can of beans is directly proportional to the height of the can. If the volume of the can is 200 in^3 when the height is 10.55 in., find the volume of a can with height 15.65 in.
Re 3): If the radii are that same then you can relate the height directly to the volume.
If the radii cannot be assumed to be exactly the same and to rather be in proportion in the same way as the height - then use the method with the squares of the radii.
Hence, it depends on the interpretation of the question which will then give an answer of V= 296.68 in^3
For number 3, one doesn't need to calculate the radii. The radii can be assumed to be equal, since the problem specifies that the volume is directly proportional to the height. Therefore:
V = kh
V = volume
k = proportionality constant
h = height
Or, k = V/h
So, using this, we can write ratios for each can:
200/10.55 and V/15.65
For this problem, these are equal"
200/10.55 = V/15.65
Solving the proportion, V = 296.68 in^3
To solve we need to use the product rule ( x +), quotient rule (`-:` - ) and power rule (the power is represented by the number in front of the log) of logarithms:
Simplify: `7 log x + 1/8 logy - 3/8 logz - log a`
`= log x^7 +log y^ (1/8) - (log z^(3/8) + log a)` (power rule)
`= (log(x^7 .y^(1/8))/(log(z^(3/8).a)))` (product rule and quotient rule)
2) log 10 000 is the same as `log 10^4 = 4` and so `log 10 000.333= 4.00001446 = 4.0000` (rounded to 4 decimals)` `
3) `V= pi r^2 h` for a cylinder. Find the radius of the first can where the height=10.55 in and the volume = 200 in^3.
`therefore 200=pi r^2 times 10.55`
and we know that the cans are directly proportional so can calculate the radius of the second can. If:
`10.55/2.46 = 4.28862`
`therefore 4.28862 = 15.65/r`
`therefore r= 3.65`
In direct proportion we can relate the squares of the radii of the two cans with the volumes:
`therefore 2.46^2/3.65^2= 200/v`
`therefore V= 440.2967 = 440.30 units^3 `
1) `log((x^7.y^(1/8))/(z^(3/8).a))` ` `
3) V= 440.30 in^3