You should use a calculator for a question like this. It is quicker and simpler.
Log(14) = 1.15
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
The expression "log(25) log(25)" represents the square of the logarithm of 25. If we let ( x = \log(25) ), then the expression simplifies to ( x^2 ). The value of ( \log(25) ) can be calculated as ( \log(5^2) = 2\log(5) ). Thus, ( \log(25) log(25) = (2\log(5))^2 = 4(\log(5))^2 ).
how do i log in
log(5)125 = log(5) 5^(3) = 3log(5) 5 = 3 (1) = 3 Remember for any log base if the coefficient is the same as the base then the answer is '1' Hence log(10)10 = 1 log(a) a = 1 et.seq., You can convert the log base '5' , to log base '10' for ease of the calculator. Log(5)125 = log(10)125/log(10)5 Hence log(5)125 = log(10) 5^(3) / log(10)5 => log(5)125 = 3log(10)5 / log(10)5 Cancel down by 'log(10)5'. Hence log(5)125 = 3 NB one of the factors of 'log' is log(a) a^(n) The index number of 'n' can be moved to be a coefficient of the 'log'. Hence log(a) a^(n) = n*log(a)a Hope that helps!!!!!
log(36,200) = 4.558709 (rounded)log[log(36,200)] = 0.658842 (rounded)
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
First, since NaOH is a base you have to find the pOH first so you use the equation -> pOH = -log[NaOH] pOH = -log[NaOH] = -log[0.0111] pOH = 1.955 Then you use this equation -> 14 = pH + pOH to find the pH 14 = pH + pOH pH = 14 - pOH = 14 - 1.955 pH = 12.045 and that makes it basic Hope that helped. ^_^
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.
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.
pH=-log[H+] pH-log(1.2x10^-3) pH=2.92 since the the pH plus the pOH is always equal 14 14-2.92=11.08 so the pOH is 11.08