log(314.25e) = log10(314.25) + log10e = 2.9316
logx(3) = log10(7) (assumed the common logarithm (base 10) for "log7") x^(logx(3)) = x^(log10(7)) 3 = x^(log10(7)) log10(3) = log10(x^(log10(7))) log10(3) = log10(7)log10(x) (log10(3)/log10(7)) = log10(x) 10^(log10(3)/log10(7)) = x
log10(225)2 equals 5.5327625985087111
Common
The natural logarithm is calculated to base e, where e is Euler's constant. For any number, x loge(x) = log10(x)/log10(e)
ln(x) = log10(X)/log10(e)
log(314.25e) = log10(314.25) + log10e = 2.9316
log10(0.083) = -1.0809 (rounded)
logx(3) = log10(7) (assumed the common logarithm (base 10) for "log7") x^(logx(3)) = x^(log10(7)) 3 = x^(log10(7)) log10(3) = log10(x^(log10(7))) log10(3) = log10(7)log10(x) (log10(3)/log10(7)) = log10(x) 10^(log10(3)/log10(7)) = x
One possible answer is log(330) Another is 1 + log(33)
Let us assume you have a Hydrochloric acid solution of 0.1 M. The pH is - log10[H+]. So log10[0.1] = -1 easy way to remember this is 103 =1000 log 101000 = 3 102 =100 log 10100 = 2 101 =10 log 1010 = 1 100 =1 log 101 = 0 10-1 =0.1 log 100.1 = -1 10-2 =0.01 log 100.01 = -2 So log10[0.1] = -1 and thus pH is - log10[H] = (minus minus 1) = 1
The little 'p' means -log10 (that's the negative log to base 10). Thus pH means -log10(Hydrogen ion concentration) → pH of the solution = -log10(7.0 x 10-2) ≈ 1.15
log5 +log2 =log(5x2)=log(10)=log10(10)=1
log10(225)2 equals 5.5327625985087111
log(6) or log10(6) = 0.778 (3sf). Therefore 100.778 = 6 (if you did not understand logarithms).
PH is equal to -log[ hydrogen ion]. You always have to use the concentration of the hydrogen ions. The concentration has to be in Molarity (moles per liter). If your concentration was actually 10 moles per liter, your PH would be -1. If it was 8 moles per liter, it would have a PH of -.903. The PH can actually go below 0 if you are using a strong acid in a very high concentration, but that would be dangerous.
The little 'p' means -log10 (that's the negative log to base 10). Thus pH means -log10(Hydrogen ion concentration) → pH of the solution = -log10(7.0 x 10-2) ≈ 1.15