The value of log16 0.0036 can be rounded to two decimal places. What is the approximate
value? (Hint: log102=0.301, log103=0.477)
(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
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.
check my answer:
16^-2.029446 = .0036
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
thank u all <3
To find the approximate value of log16 0.0036, we can use the property of logarithms that states:
log a b = log c b / log c a
First, let's rewrite the expression: log16 0.0036 = log10 0.0036 / log10 16
Using the given hints, we can approximate log10 0.0036 and log10 16 as follows:
log10 0.0036 = log10 3.6 × 10^(-3) = log10 3.6 - log10 10^3 = log10 3.6 - 3
From the hints, we know that log10 3.6 is approximately 0.477. Thus:
log10 0.0036 ≈ 0.477 - 3 ≈ -2.523
Now, let's substitute the values back into the original expression:
log16 0.0036 = log10 0.0036 / log10 16 ≈ -2.523 / 0.301
Calculating this division, we find that:
log16 0.0036 ≈ -8.40
Rounded to two decimal places, the approximate value of log16 0.0036 is -8.40.