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How do you figure time on simple interest?
p = principal r= rate in decimal a =amount or final total t = time in years i = accumulated interest Simple interest: i=prt therefore t = i/pr Compound interest: a=(1+r)^t alright I'm going to throw in logs (logarithms) baby therefore log a = log ((1+r)^t)=t x log (1+r) therefore t = log a / log (1+r)
What is log base 4 of 27?
You can calculate that on any scientific calculator - like the calculator on Windows (if you change the options, to display as a scientific calculator). Log base 4 of 27 is the same as log 27 / log 4. You can use logarithms in any base to calculate that - just use the same base for both logarithms.
How do you solve log base16 8?
You divide log 8 / log 16. Calculate the logarithm in any base, but use the same base for both - for example, ln 8 / ln 16.
How do you calculate delta t?
To calculate Delta t, you would subtract Universal Time or UT from Terrestrial Time or TT. Delta t would be the difference.
How do you calculate the exponent y if the base is 2 and the answer is 50?
If 2y = 50 then y*log(2) = log(50) so that y = log(50)/log(2) = 5.6439 (approx). NB: The logarithms can be taken to any base >1.
How do you calculate Mean Generation Time?
g=(log Nt- log Nto)/log 2 where N=absorbance reading @ time indicated MGT= (t-to)/g
How do you solve a log?
You calculate a log, you do not solve a log!
How do you calculate a log with 9 diameter and 10' length?
You do not calculate a log!You can calculate the surface area or the volume or, if you know the species, the mass or even time for which it would burn in a hearth. But the log, itself, is not something you can calculate.You do not calculate a log!You can calculate the surface area or the volume or, if you know the species, the mass or even time for which it would burn in a hearth. But the log, itself, is not something you can calculate.You do not calculate a log!You can calculate the surface area or the volume or, if you know the species, the mass or even time for which it would burn in a hearth. But the log, itself, is not something you can calculate.You do not calculate a log!You can calculate the surface area or the volume or, if you know the species, the mass or even time for which it would burn in a hearth. But the log, itself, is not something you can calculate.
What is T if two and a half billion equals one hundred forty times-doneseparately-three-fifths to the power of t?
2,500,000,000 = 140*(3/5)t Therefore (3/5)t = 2,500,000,000/140 => t*log(0.6) = log(2,500,000,000/140) => t = log(2,500,000,000/140)/log(0.6) = -32.6881
How do you make T the subject of this equation X is equal to W times A to the power of T?
X=W*(A)^T Use logarithms. T=log(A)/log(X/W)
How do you figure time on simple interest?
p = principal r= rate in decimal a =amount or final total t = time in years i = accumulated interest Simple interest: i=prt therefore t = i/pr Compound interest: a=(1+r)^t alright I'm going to throw in logs (logarithms) baby therefore log a = log ((1+r)^t)=t x log (1+r) therefore t = log a / log (1+r)
How do you calculate GPA for Indian students?
log on to ece.org
How you calculate the acidity value?
pH = - log[H3O+]
How do you calculate Log2 in Excel?
Use the LOG function. =LOG(n,b) n = Number b = Base =LOG(2,10) = 0.30103
What is the relationship between absorbance and transmittance?
Absorbance = -log (percent transmittance/100)
What is t pains website?
log on to tpain
Integration of tangent cubed of x?
Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x. * * * Here is how we can obtain this result: First, let t = tan x, s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx, and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ = t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d log c. Combining these results, we have dI = t dt - t dx = ½ d(t²) - d log c. This integrates readily, giving I = ½ t² - log c, which is the solution we sought. * * * We may check our result, by differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus, (d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc = 1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s / c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²) - t = t + t³ - t = t³, re-assuring us that we have integrated correctly.
