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How would you solve 250 to the x power equals 400000?
250(x) = 400,000 Use logs to base '10' Hence log250^(x) = log400,000 xlog250 = log 400,000 Notirce how the power (x) becomes the coefficient. This is oerfectly correct under 'log' rules. x = log 400,000 / log 250 (NOT log(250/400,000). x = 5.60206 / 2.39794 x = 2.33619689..... ~ 2.33619 to 5 d.p.
What is the answer to cot squared x - tan squared x equals 0?
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
What is log to the third power equal to?
Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.
Rules of log?
Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0
Log9x plus log x equals 4log10?
log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3
Sec x times sin x divided by tan x?
1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1
What is a constant coefficient?
In a ploynomial or differential equation or really any formula or equation with variables in it, the coefficients are the terms "in front of" the variable or multiplied the variables. Each variable generally has its own coefficient. If a coefficient is constant (ie just a number) then it is a constant coefficient. eg Consider the polynomial , 3x2+9yx+6 in terms of x. It has one constant coefficient (3), one variable coefficient (9y) and one constant (6).
What is the derivative of ln cos x?
The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
What does cos divided by sin equal?
tan x
What is log to the third power equal to?
Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.
Log x - log 6 equals log 15?
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
Rules of log?
Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0
Log x plus log 2 equals log 2?
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
Log9x plus log x equals 4log10?
log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3
What is the derivative of tan x?
The derivative of tan(x) is sec2(x).(Which is the same as 1/cos2(x).
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
