You should use a calculator for a question like this. It is quicker and simpler.
Log(14) = 1.15
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∙ 13y agolog(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
how do i log in
log(36,200) = 4.558709 (rounded)log[log(36,200)] = 0.658842 (rounded)
False When logs are taken, division becomes subtraction, so the log of a quotient is the log of the numerator minus the log of the denominator.
Assuming you are asking about the natural logarithms (base e):log (-1) = i x pithereforelog (log -1) = log (i x pi) = log i + log pi = (pi/2)i + log pi which is approximately 1.14472989 + 1.57079633 i
You will need 14 two's multiplied together to equal 16384. the answer to this can be found by log2(16384) = 14. Since most calculators don't have log base 2, you can do this: log(16384)/log(2) = 14. You can use the 'base 10' log or natural log [ln(16384)/ln(2) = 14]
Adam-12 - 1968 Log 14 SWAT 2-14 was released on: USA: 24 January 1970
0.03
The Louise Log - 2007 How to Fake It 2-14 was released on: USA: 1 April 2012
To find the pH of a 0.002614M NaOH solution, first find the pOH by taking the negative base-10 logarithm of the hydroxide ion concentration. pOH = -log(0.002614) = 2.583. Since pH + pOH = 14 for water at 25°C, pH = 14 - pOH = 14 - 2.583 = 11.417.
Navy Log - 1955 The Beach Pounders 3-14 was released on: USA: 19 December 1957
It depends on what compound is dissolved in what solvent. Presuming a strong monoprotic acid (eg. HCl) in water the pH is equal to -log[H+] = -log(6.8*10^-3) = 3-log6.8 = (3-0.8) = 2.2 Presuming a strong monoprotic base (eg. NaOH) in water the pH is equal to 14 - pOH = 14 - (-log[OH-]) = 14+log(OH-) = 14 + log(6.8*10^-3) = 14 + (-3 + log6.8) = 14 + (-3+0.8) = 14 - 2.2 = 11.8 Fot pH calculation of weak acids and bases you'll need more complex formula and approximations.
MacGyver - 1985 Log Jam 5-14 was released on: USA: 5 February 1990 Germany: 12 March 1991
Adam-12 - 1968 Log 115 Gang War 3-14 was released on: USA: 9 January 1971
The pOH is 6,4.
[OH-] = 1x10^-3 M[H+][OH-] = 1x10^-14[H+] = 1x10^-14/1x10^-3 = 1x10^-11pH = -log 1x10^-11 = 11Done another way:pOH = -log [OH-] = -log 1x10^-3 = 3pH + pOH = 14pH = 14 - 3 = 11
For a 0.1 M HCl solution, all of the HCl molecules will dissociate into H+ and Cl- ions, resulting in a concentration of 0.1 M for H+ ions. Since pH = -log[H+], the pH of the solution is 1. The pOH can be calculated by subtracting the pH from 14, so the pOH would be 13.