The value of log16 0.0036 can be rounded to two decimal places. What is the approximate

Question

value? (Hint: log102=0.301, log103=0.477)

Answer 1

(Hint: log102=0.301, log103=0.477)
maybe
(Hint: log10 (2) =0.301, log10 (3)=0.477)

16^ log16 (0.0036) = 0.0036
16^x = 0.0036
x log10 (16) = 0.0036
x log10 (2^4) = 0.0036
4 x log10 (2) = 0.0036
4* 0.301 x = 0.0036
x = 0.0036 /1.204
= 0.00299 = 0.003

Answer 2

Fixing your typing:
The value of log₁₆ 0.0036 can be rounded to two decimal places. What is the approximate
value? (Hint: log 2=0.301, log 3=0.477)

(if your base of the log is 10, you don't have to write it, it is understood to be 10)

let x = log₁₆ 0.0036
then 16^x = 0.0036
2^(4x) = 36 * 10^-4
= 3*3*2*2 * 10^-4
= 3^2 * 2^2 * 10^-4

2^(4x-2) = 3^2 * 10^-4
take log of both sides, and use your rules of logs
(4x-2) log2 = 2log3 - 4log10
(4x-2) log2 = 2log3 - 4 , since log10 = 1
4x - 2 = (2log3 - 4)/log2
4x - 2 = (2(.477) - 4)/.301
4x - 2 = -10.1196
4x = -8.1196
x = -2.03

Question: Since we both would have used a calculator to do the last few
steps, why not just do:
log .0036/log 16 = -2.029446 , which is -2.03 rounded to 2 decimals.

Answer 3

check my answer:

16^-2.029446 = .0036

Answer 4

whoops !
16^x = 0.0036
x log10 (16) = log10 (0.0036) forgot to take log on right
4 x log10 (2) = log10(0.0036)
4 x * 0.301 = -2.44
x = -2.03

Answer 5

thank u all <3