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Find the exact solutions, in their simplest form, to the equations : a) 2ln(2x + 1)-4=0 b)7^(x)e^(4x)=e^5

a) 2ln(2x + 1)-4=0

-> ln(2x + 1)-2=0

-> ln(2x + 1)=2

-> (2x + 1)=e²

-> 2x = e²-1

-> x = (e²-1)/2

b) 7^x

Answered by Elizabeth L. • Maths tutor

4589 Views

How do I find the equation of the tangent to y = e^(x^2) at the point x = 4?

This question can be broken up into two main parts, one you're probably familiar with from C1 + C2 and one which is newer. The first part relates to differentiating a compound function (where a function i...

Answered by Sam H. • Maths tutor

9128 Views

differentiate with respect to 'x' : ln(x^2 + 3x + 5)

(2x + 3) / (x² + 3x + 5)

Answered by Alec S. • Maths tutor

4772 Views

Differentiate with respect to x: y=(6x^2-1)/2sqrt(x)

1st Step: Must separate the fractions and simplify:

y= (6x²-1)/2sqrt(x)= 6x²/(2x^1/2) - 1/(2x^1/2)

Cancel 6 with 2 and x

Answered by Elena S. • Maths tutor

8641 Views

i) It is given that f(x)=(-5-33x)/((1+x)(1+5x)), express f(x) in the form A/(1+x) + B/(1+5x) where A,B are integers. ii) hence express the integral of f(x) between x=3 and x=0 in the form (p/q)ln4 where p,q are integers.
i) -7/(1+x) + 2/(1+5x) ii) (-31/5)ln4

DT
Answered by Damon T. • Maths tutor
4029 Views

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Top answers

Find the exact solutions, in their simplest form, to the equations : a) 2ln(2x + 1)-4=0 b)7^(x)e^(4x)=e^5

How do I find the equation of the tangent to y = e^(x^2) at the point x = 4?

differentiate with respect to 'x' : ln(x^2 + 3x + 5)

Differentiate with respect to x: y=(6x^2-1)/2sqrt(x)

i) It is given that f(x)=(-5-33x)/((1+x)(1+5x)), express f(x) in the form A/(1+x) + B/(1+5x) where A,B are integers. ii) hence express the integral of f(x) between x=3 and x=0 in the form (p/q)ln4 where p,q are integers.

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