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MathRewrite as `log_e ` `log_(e)x=1 ` `e^1=x ` `x=0.368 `

MathAssume `log_10 ` `log_(10)x=2 ` `10^2=x ` `x=0.01 `

MathRewrite the given equation ` 4^3=x` `x=64 `

MathRewrite the given equation `5^(1/2)=x ` `x=2.236 `

MathSince the bases are the same, set the exponents equal to each other `x=x^22 ` `0= x^2x2 ` `0=(x2)(x+1) ` `x=2 ` and `x=1 `

MathSince the bases are the same, set the exponents equal to each other `x^23=x2 ` `x^2x1=0 ` Use the quadratic formula `x=0.618 ` and `x=1.618 `

MathDivide both sides by 4 `3^x=5 ` Take the ln of both sides `ln3^x=ln5 ` Bring down the exponent `xln3=ln5 ` `x=ln5/ln3 ` `x=1.465 `

MathDivide both sides by 4 `e^x=91/4 ` Take the ln of both sides and bring down the exponent `xlne=ln(91/4) ` `x=ln(91/4)/lne ` `x=3.125 `

MathAdd 9 to both sides `e^x=28 ` Take the ln of both sides and bring down the exponent `xlne=ln28 ` `x=ln28/lne ` `x=3.332 `

MathSubtract 10 from both sides `6^x=37 ` Take the ln of both sides `ln(6^x)=ln37 ` Bring down the exponent `xln6=ln37 ` `x=ln37/ln6 ` `x=2.015 `

MathTake the ln of both sides `ln(3^(2x))=ln80 ` ` ` Bring down the exponent `2xln3=ln80 ` `x=ln80/(2ln3) ` `x=1.994 `

MathTake the ln of both sides `ln(4^(3t))=ln0.10 ` Bring down the exponent `3tln4=ln0.10 ` `t=ln0.10/(3ln4) ` `t=0.554 `

MathTake the ln of both sides `ln(2^(3x))=ln565 ` Bring down the exponent ` (3x)ln2=ln565` `3x=ln565/ln2 ` `x=ln565/ln23 ` ` x=6.142`

MathTake the ln of both sides `ln(8^(2x))=ln431 ` Bring down the exponent `(2x)ln8=ln431 ` `2x=ln431/ln8 ` `x=ln431/ln8+2 ` `x=4.917 `

MathDivide both sides by 8 `10^(3x)=1.5 ` Take the ln of both sides `ln(10^(3x))=ln1.5 ` Bring down the ` 3x` by log rules `3xln(10)=ln1.5 ` `x=(ln1.5)/(3(ln(10))) ` `x=0.059 `

MathGet the base to be the same `4^(x)=4^(2) ` Therefore, `x=2 `

MathAnother method to solve this would be to get the same base because there is a rule that states if the bases are the same, then the exponent must equal each other. Given, `x^a=x^b ` Then, `a=b `...

Math`ln(x)=ln(2) ` Therefore, `x=2 `

Math(a) Take the derivative. Use the product rule. `y'=(x2)(2x+3)+(x^2+3x)(1)= (x2)(2x+3)+(x^2+3x)` Given the point `(1,4)` , substitute `x=1` . `y'(1)= (12)(2(1)+3)+(1^2+3(1)) =...

MathSo I will start by rewriting this equation in the form: `f(x)=x^42x^2+3 ` We know that if we have then we can plug in any value for x, and the result will be what the slope of the function is at...

MathTake the derivative of `y=x^3+x` . `y'= 3x^2+1` For horizontal tangent lines, the slope is zero. Set the derivative function to zero because we want to know where x is when the slope is zero....

MathRewrite the equation and take the derivative. Use the power rule. `y= 1/x^2 = x^(2)` `y' = 2x^(3)` `y' = 2/x^3` For horizontal tangent lines, the slope will always equal zero. Set the...

MathSo I will start by rewriting this equation in the form: `f(x)=x^2+9 ` We know that if we have `f'(x)` then we can plug in any value for x, and the result will be what the slope of the function is...

MathWe will need to take the derivative of the function and set the derivative equal to zero. `y' = 1+cos(x)` `0 = 1+cos(x)` `cos(x)= 1` The value of `x` is the angle where we have a point on the unit...

MathTake the derivative of the function. `y' = sqrt3 2sin(x)` Set the derivative function equal to zero, since we want to know where x is when the slope is zero. `0 = sqrt3 2sin(x)` `2sin(x)=sqrt3`...

MathThe tangent line will touch a point on the function f(x). Set both equations equal to each other. `kx^2= 6x+1` With two unknown variables, we will need another relationship of x and k. Take the...

MathThe tangent line will touch a point on the original function f(x). Set the two equations equal to each other since they intersect. `kx^2=2x+3` We will need another relationship since we have two...

MathThe original equation will intersect a point on the tangent line. Set both equations equal to each other. `k/x = 3/4 x +3` We have 2 unknowns and 1 equation. Take the derivative of f(x) in order...

MathThe tangent line will touch one point of the original function. Set the two equations equal to each other. Rewrite the square root as a fractional power. `kx^(1/2)= x+4` We need a relationship of...

MathThe tangent line must touch a point on the f(x) function. Set both equations equal to each other. `kx^3 = x+1` Take the derivative of f(x) to find another relationship with k and x. The k is a...

MathThe tangent line will touch one point of the original function. This means that: `kx^4 = 4x1` We have two variables that we don't know. Find the derivative of `f(x)` . The k is a constant and...

Mathg(t) = 2*cost + 5 slope of the curve = g'(t) = d{g(t)}/dt = 2*sint Thus, slope at (pi,7) g'(pi) = 2*sin(pi) =0 Thus, the curve has a tangent parallel to the xaxis(horizontal axis) at (pi,7)

MathThe derivative of the first term can be solved using the power rule. `d/dx x^n = nx^(n1)` The derivative of `x^2` is `2x` . The derivative of a constant is always zero. Use the power rule again...

MathUse the power rule for each term. Derive each term. `d/dx x^n= nx^(n1)` `f'(x) = 3x^2 2 9x^4` Eliminate all negative exponents. The answer is: `f'(x) = 3x^2 2 9/x^4`

MathWe can rewrite g(t) so that we can avoid the quotient rule altogether. `g(t)=t^24t^3` Use the power rule to derive g(t). The power rule is: `d/dx x^n = nx^(n1)` `g'(t)= 2t +12 t^4` The answer...

MathDerive term by term. The derivative is the slope of a function. The derivative of `8x` is 8. Rewrite `3/x^2` with a negative exponent. `3/x^2= 3x^2` Use the power rule to derive. `2(3)x^3...

MathUse the quotient rule to solve this question. The formula is: `f'(x) = (BT'  TB')/ B^2` Where B and T are bottom function and top function, respectively. `B= x` `B'=1` `T=4x^3+3x^2` `T'=...

MathDivide each term by `x^3` . `f(x) = 2xx^(2)` Use the constant rule and the power rule to derive this function. `f'(x) = 2+2x^3` Eliminate the negative exponent by rewriting it as a fraction. The...

MathDivide each term by to avoid the quotient rule. `f(x) = x3+4x^2` Take the derivative of each term. Use the power rule for the last term. `f'(x)= 10+(2)(4)(x^3)` `f'(x) = 18x^3` Rewrite the...

MathDivide all terms by x. `f(x)=4x^2+2+5/x ` `f(x)= 4x^2+2+5x^1` Take the derivative of each term. `f'(x) = 8x+0 +(1)(5)(x^2)` Simplify. The answer is: `f'(x) = 8x5/x^2` ` `

MathGiven: ` y=x(x^2+1)` `y=x^3+x` `y'=3x^2+1` ``

MathGiven: `y=x^2(2x^23x)` Distribute in the x^2 to rewrite the expression as a polynomial. `y=2x^43x^3` Find the derivative. `y'=8x^39x^2`

MathRewrite the function with exponents. `f(x) = x^(1/2)6x^(1/3)` Use the power rule to solve. `f'(x) = 1/2 x^(1/2)(1/3)6x^(2/3)` Simplify and eliminate the negative fractions by writing them as a...

MathGiven: `f(t)=t^(2/3)t^(1/3)+4` The derivative is: `f'(t)=(2)/(3)t^(1/3)(1)/(3)t^(2/3)` `f'(t)=(2)/(3root(3)(t))(1)/(3root(3)(t^2))`

MathRewrite the function: `f(x)=6x^(1/2)+5cos(x)` Take the derivative of the first term by using the power rule and the derivative of the second term. `f'(x)= 1/2 (6)x^(1/2)5sin(x)` Simplify the...

MathThe first term is the same as: `2/x^(1/3) =2x^(1/3)` Take the derivative of this term. `1/3(2)x^(1/31) = 2/3x^(4/3)` The derivative of `3cos(x)` is `3sin(x)` . Combine the terms....

MathGiven: Take the derivative of this function. `y' = 4x^36x` Substitute the value of x=1 from the given point into the derivative function. `y' = 4(1)^36(1)` `y'=46 = 2` The slope at the...

Matha) So we are given: `y=x^3  3x` If we derive this function, then we will have the following: `y'=3x^2 3` This will show us what the slope of the original line would be at any x value that we plug...

MathEliminate the radical by rewriting it as a fraction. The function becomes: `f(x) = 2/ (x^3)^(1/4) = 2/x^(3/4) = 2x^(3/4)` Take the derivative by using the power rule. `f'(x) = 3/4...

MathGiven: `f(t)=2t^2+3t6` The derivative is: `f'(t)=4t+3` ``