My teacher assigned us this problem:
Using the following logarithms
Log2,log3 and log10 solve for the log150,log 75 incorporating the properties of logarithms ,i tried ,but failed plz help!!
Well, with log150 it's log3+log(10)^2-log2
With log75 it is equivalent to log3+log(10)^2-log(2)^2
Hope this helped
not quite. log(x^2) = 2logx
150 = 3*100/2, so
log 150 = log3 + 2log10 - log2
75 = 3*(10/2)^2, so
log75 = log3 + 2log10 - 2log2
or, since 75 = 150/2
log75 = log150-log2 as above
To solve for log150 and log75 using the given logarithms (log2, log3, and log10) and the properties of logarithms, you can use the following steps:
1. Start by expressing the given numbers, 150 and 75, as powers of the bases 2, 3, and 10. This will help us apply the properties of logarithms.
For example:
- 150 can be expressed as 2^1 * 3^1 * 5^2 (since 5 * 5 = 25 and 25 * 6 = 150)
- 75 can be expressed as 3^1 * 5^2 (since 3 * 25 = 75)
2. Now, let's use the property of logarithms that states: log(A * B) = log(A) + log(B).
Applying this property:
- log150 = log(2^1 * 3^1 * 5^2)
= log(2^1) + log(3^1) + log(5^2)
= log2 + log3 + 2*log5
- log75 = log(3^1 * 5^2)
= log(3^1) + log(5^2)
= log3 + 2*log5
3. To solve for log2, log3, and log5, we can use the properties of logarithms.
For example:
- log2 can be approximated as 0.301
- log3 can be approximated as 0.477
- log5 can be approximated as 0.699
4. Now, substitute the approximations into the expressions for log150 and log75:
- log150 ≈ 0.301 + 0.477 + 2*0.699
- log75 ≈ 0.477 + 2*0.699
5. Use a calculator or calculator software to compute the approximations to get the final values for log150 and log75.
That's it! By expressing the given numbers in terms of appropriate powers of the bases 2, 3, and 10, and applying the properties of logarithms, you can find the approximate values of log150 and log75.