What is the value of #x# where the tangent to #y -1 = 3^x# has a slope of #5#?

Question

What is the value of #x# where the tangent to  #y -1 = 3^x# has a slope of #5#?

Emily Ammerman · Accepted Answer

#x= ln(5/ln3)/ln3 ~=1.379#

Start by finding the derivative of the function.

#y - 1 = 3^x#

#ln(y - 1) = ln(3^x)#

#ln(y - 1) = xln3#

Differentiate the left hand side using the chain rule and the right hand side using the product rule.

#1/(y - 1) xx1(dy/dx) = 1(ln3) + 0(x)#

#1/(y- 1)dy/dx= ln3#

#dy/dx = ln3/(1/(y - 1))#

#dy/dx= ln3(y - 1)#

#dy/dx= ln3(3^x + 1 - 1)#

#dy/dx= 3^xln3#

The derivative represents the instantaneous rate of change of a function at any given point #x= a# within the domain of the function. In this problem, we want to find where the tangent line is parallel to #y = 5x- 1#. The line #y =5x - 1# has a slope of #5#, parallel lines have equal slopes, so we will want to find where on the derivative the #dy/dx = 5#.

#5 = 3^xln3#

#5/ln3 = 3^x#

#ln(5/ln3) = xln3#

#ln(5/ln3)/ln3 = x#

If you would like an approximation, use a calculator to obtain #x = 1.379#.

Hopefully this helps!

Ezra Adams · Answer

undefined To find the value of x where the tangent to y - 1 = 3^x has a slope of 5, we need to differentiate the equation with respect to x. The derivative of y - 1 = 3^x is dy/dx = 3^x * ln(3).

To find the slope of the tangent line, we set dy/dx equal to 5 and solve for x:

5 = 3^x * ln(3)

Divide both sides by ln(3):

5/ln(3) = 3^x

Take the logarithm of both sides with base 3:

log3(5/ln(3)) = x

Therefore, the value of x where the tangent to y - 1 = 3^x has a slope of 5 is x = log3(5/ln(3)).