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Logarithm questions for Algebra 2/Trig?
The numbers in parenthesis are the bases.
1. log(2)32 - log (2)128
2. log(1/2)2 + log (1/2) 2^1/2
3. Use properties of logarithms with the fact that log2 = 0.301 to evaluate.
a) log20 b) log200
4. Most swimming pool experts recommend a pH of 7.0 and 7.6 for water in a swimming pool. Use pH = -log[H+], and write an expression for the difference in hydrogen ion concentration in this pH range.
Please show your work and explain how you got your answers. I have a quiz tomorrow and need to bring my grade up.
And thanks for any help!
1 Answer
- james brownLv 49 years agoFavorite Answer
Use log and exponent laws. If two logs of same base are being added, you make that into one log and the term in brackets will be the product of the two terms, and if they were being substracted, then it will be the first term divided by 2nd term. Also log 1/2 (2)= -1 and log2 (1/2)= -1 those are just ratios, so if you can turn the term in brackets to something of the powers of the base, do it cuz it simplifies a lot, you can just read off the exponent and that will be your answer like i do in the first two problems..
1) log2 (32/128)= log 2 (1/4)= log 2 (2^-2)= -2
2) log1/2 (2*2^1/2)= log 1/2 (2^3/2)= log 1/2 ((1/2)^-3/2)= -3/2
3)a) log 20= log (2*10)= log 2 + log 10= 0.301 + 1= 1.301
b) log 200= log (2*100)= log 2 + log 100= 0.301 + 2= 2.301
4) ph = -log [h+] , 10^-ph= [ H+]
so change in [H+]= 10^-7 - 10^-7.6
