What is the correct option from given options?
pls explain your answer
↳Redirected from
"What are valence electrons?"
Answer nº 2
#1 + 2log_10 2#
Let's start by putting 250 in the form of 25*10:
#" " log_10 (25*10)#
Using the multiplication property of logarithms (#log_a (x*y) = log_a x + log_a y#); and writing 25 as #5^2#
#" " log_10 (5^2*10) = log_10 10 + log_10 5^2#
Knowing that #log_a a = 1# (#log_10 10 = 1#), and the power property of logarithms (#log_a x^n = n log_a x#)
#" "log_10 10 + log_10 5^2 = 1 + 2 log_10 5#
Hope that helped!!
Option 3) #3-2log_(10) 2#
There are a few ways to do this, but this is the method I used. It will use the following properties of logarithms:
#log_a (b^c) = c*log_a b color(white)(aaaaaa)"Exponent Property"#
#log_a (b/c) = log_a b - log_a c color(white)(aaaaaa)"Quotient Property"#
#log_a a = 1 color(white)(aaaaaaa)"Identity Property"#
Begin by recognizing that #250 = 1000/4#, and rewrite the original logarithm:
#log_10 250 = log_10 (1000/4)#
Now apply the Quotient Property above to split this logarithm into two separate logarithms:
#log_10 (1000/4) = log_10 1000 - log_10 4#
Rewrite both logarithms using exponents:
#log_10 1000 - log_10 4 = log_10 (10^3) - log_10 (2^2)#
Apply the Exponent Property:
#log_10 (10^3) - log_10 (2^2) = 3*log_10 10 - 2*log_10 2#
Finally, apply the Identity Property:
#3*log_10 10 - 2*log_10 2 = 3(1) - 2*log_10 2 = 3 - 2 log_10 2#
