I assume you are asking how to solve a logarithmic equation. Well let's quickly review what the log function is: for the equation log(x)=y, we are saying that 10^y=x.
So once you have isolated the logarithm, take the value of the base, raise it to the nth power (when 'n' is the value that the function is equal to) and set that equal to the value inside of the log.
log(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(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)
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.
please answer thee please....................
it is not affected by fuse. the ECU controls the idle electronically. you need to do an idle relearn prosedure.
log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)
tom dunsdons dad and mum log log log log log log log in my buttt
Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."
For a quotient x/y , then its log is logx - log y . NOT log(x/y)
"Log" is not a normal variable, it stands for the logarithm function.log (a.b)=log a+log blog(a/b)=log a-log blog (a)^n= n log a
log on to
1
log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90
No. log 20 is a positive number , so it you subtract it from log 5 you get less than log 5. However, log10 5 = 1 - log102 = 2- log1020 . or log 5 - log 20 = log 5 - log 4*5 = log 5 - (log 5 + log 4) = log 5 - log 5 - log 4 = - log 4 But we do not need to do all of these computations, because log 5 is different from log 5 - log 20 by the law of the equality that says two equals remain equal if and only if we subtract (in our case) the same thing from them.
You have to use logarithms (logs).Here are a few handy tools:If [ C = D ], then [ log(C) = log(D) ]log(AB) = log(A) + log(B)log(A/B) = log(A) - log(B)log(Np) = p times log(N)