
How do you simplify #Log(x+4)=2log(x-2)#? - Socratic
Dec 25, 2015 · #x+4=(x-2)^2# Distribute and simplify. #x+4=x^2-4x+4# #0=x^2-5x# #0=x(x-5)# #x=0,5# However, the answer #x=0# is thrown out, since plugging in #0# in #2log(x-2)# would …
2·log(5x - 4) - log 4 = log (x - Socratic
Mar 2, 2018 · However, for #x=0#, in the original equation, we end up with a #log(-4)# in the equation. Because, in the domain of real numbers, we are not allowed to have logarithms of …
How do you solve 2log_5x = log_5 9? - Socratic
Feb 24, 2016 · First simplify using the rule alogn = logn^a log_5(x^2) = log_5(9) Put everything to one side of the equation. 0 = log_5(9) - log_5(x^2) Simplify using the rule log_am - log_an = …
How do you solve log_8 25= 2log_8 x? - Socratic
Apr 11, 2016 · x = 5 Rearrange using laws of logarithms, where a coefficient outside a log is the same as a power inside it. log_8 25 = 2log_8 x = log_8 x^2 Raise both sides by eight, 8^(log_8 …
2log x-log 5 =-2? - Socratic
May 21, 2018 · x=sqrt5/10 As 2logx-log5=-2 2logx=-2log10+log5 or logx^2=log(5/10^2) or x^2=5/10^2 or x=sqrt5/10 Observe that though there could have been an answer x=-sqrt5/10 …
How do you solve 2log_2x-log_2 5=2^2? - Socratic
Apr 10, 2016 · x=4sqrt5 As log_ba=loga/logb 2log_2 x-log_2 5=2^2 can be written as 2logx/log2-log5/log2=4 and multiplying each side by log2 as log2!=0, we get 2logx-log5=4log2 or …
How do you differentiate #log(sin^2(x))#? - Socratic
Feb 10, 2017 · 2cot(x) This answer assumes that log(x)=log_e(x), the natural logarithm. First use the rule log(a^b)=blog(a): y=log(sin^2(x))=2log(sin(x)) From here, we need to know that …
How do you solve #log(x + 10) - log(x) = 2log(5)#? - Socratic
Sep 11, 2015 · x = 5/12 log(x + 10) - logx = 2log5 Our first step is to rewrite the equation using some laws of logarithms, specifically, logA - logB = log (A/B). This law allows us to rewrite the …
How do you solve log(x+2)+log(x-2)=1? - Socratic
Feb 25, 2016 · x=sqrt14 Simplify the left hand side through the logarithm rule: log(a)+log(b)=log(ab) Thus, we obtain log[(x+2)(x-2)]=1 Distributed, this gives log(x^2-4)=1 …
Log2 (x+1)=log4 (x^2-x+4) What is the value of x? - Socratic
Apr 18, 2018 · #2log_2(x+1)=log_2(x^2-x+4)# Now we can use the property of logarithms that says #alogx=logx^a#. This lets us rewrite our equation as. #log_2(x+1)^2=log_2(x^2-x+4)#. If …