t = 13.67. This is, at first, a tricky question. However, once you learn, this type of question is easy to do. I will divide it into steps: We start with 2=[1+(0.052/1)]^t. We will use logarithms to solve this. We'll put …

Now, notice what happens when color (white) (a) ul ( ["A"^ (-)] = ["HA"]) In this case, you have ( ["A"^ (-)])/ ( ["HA"]) = 1" " and " " log ( ( ["A"^ (-)])/ ( ["HA"])) = log (1) = 0 This will get you "pH" = "p"K_a The …

"Old pH = 4.72"; "New pH = 4.70" > The strategy to follow is a) Write the chemical equation for the buffer. b) Calculate the "pH" of the buffer. c) Calculate the moles of "HCl" added. d) Calculate the …

In most cases we know what these are so l (beta|x,y) = 1/ (sqrt (2pisigma^2))e^ (- (y-xbeta)^2/ (2sigma^2)) now we take the log to simplify thus log (l (beta|x,y)) = log (1/ (sqrt (2pisigma^2))) + (- (y …

Apr 12, 2017 · Explanation: Use the property #log (a)+log (b)=log (ab)# to simplify the equation to #log ( (x+1)* (x-1))=log (24)#.

The change of base law was also used, because I personally don't always remember the derivative of #log_10 (x)# right away, but remembering it for #lnx# is somewhat easier.

A: 50 ml B:pH=2.87 C:pH=4.74 D:pH=8.72 A: 50 ml HAc 0.1M = 5mMol HAc (weak acid) 50 NaOH 0.1 M = 5 mMol NaOH (strong base) Equivalence: 1:1, 5mMol:5 mMol => 50ml 0,.1 M NaOH B: pH = …

The solution is : x=e vv x=1/e The logarythms have different bases, so first you have to change base using formula: log_ax=log_ba/log_bx In this case we get: log_ex-1/log_ex=0 (log_ex)^2-1=0 (log_ex …

May 9, 2018 · We know that log_bx=y is a continous function defined in (0,oo) We know also that log_bx is negative in (0,1) In our case 2x-1 must be smaller than 1 but bigger than 0 (by prior paragraph).

I'm unsure if this is what the question is asking for, but we can also bring the #1# into the logaritm. Assuming that #log# means #log_10#, we can rewrite the #1# like so: