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Calculus Examples
ln(5)+12⋅ln(x+3)-3ln(1+√x)ln(5)+12⋅ln(x+3)−3ln(1+√x)
Step 1
Step 1.1
Simplify 12ln(x+3)12ln(x+3) by moving 1212 inside the logarithm.
ln(5)+ln((x+3)12)-3ln(1+√x)ln(5)+ln((x+3)12)−3ln(1+√x)
Step 1.2
Simplify -3ln(1+√x)−3ln(1+√x) by moving 33 inside the logarithm.
ln(5)+ln((x+3)12)-ln((1+√x)3)ln(5)+ln((x+3)12)−ln((1+√x)3)
ln(5)+ln((x+3)12)-ln((1+√x)3)ln(5)+ln((x+3)12)−ln((1+√x)3)
Step 2
Use the product property of logarithms, logb(x)+logb(y)=logb(xy)logb(x)+logb(y)=logb(xy).
ln(5(x+3)12)-ln((1+√x)3)
Step 3
Use the quotient property of logarithms, logb(x)-logb(y)=logb(xy).
ln(5(x+3)12(1+√x)3)
Step 4
Use the Binomial Theorem.
ln(5(x+3)1213+3⋅12√x+3⋅1√x2+√x3)
Step 5
Step 5.1
One to any power is one.
ln(5(x+3)121+3⋅12√x+3⋅1√x2+√x3)
Step 5.2
One to any power is one.
ln(5(x+3)121+3⋅1√x+3⋅1√x2+√x3)
Step 5.3
Multiply 3 by 1.
ln(5(x+3)121+3√x+3⋅1√x2+√x3)
Step 5.4
Multiply 3 by 1.
ln(5(x+3)121+3√x+3√x2+√x3)
Step 5.5
Rewrite √x2 as x.
Step 5.5.1
Use n√ax=axn to rewrite √x as x12.
ln(5(x+3)121+3√x+3(x12)2+√x3)
Step 5.5.2
Apply the power rule and multiply exponents, (am)n=amn.
ln(5(x+3)121+3√x+3x12⋅2+√x3)
Step 5.5.3
Combine 12 and 2.
ln(5(x+3)121+3√x+3x22+√x3)
Step 5.5.4
Cancel the common factor of 2.
Step 5.5.4.1
Cancel the common factor.
ln(5(x+3)121+3√x+3x22+√x3)
Step 5.5.4.2
Rewrite the expression.
ln(5(x+3)121+3√x+3x1+√x3)
ln(5(x+3)121+3√x+3x1+√x3)
Step 5.5.5
Simplify.
ln(5(x+3)121+3√x+3x+√x3)
ln(5(x+3)121+3√x+3x+√x3)
Step 5.6
Rewrite √x3 as √x3.
ln(5(x+3)121+3√x+3x+√x3)
Step 5.7
Factor out x2.
ln(5(x+3)121+3√x+3x+√x2x)
Step 5.8
Pull terms out from under the radical.
ln(5(x+3)121+3√x+3x+x√x)
ln(5(x+3)121+3√x+3x+x√x)
Step 6
Step 6.1
Use n√ax=axn to rewrite √x as x12.
ln(5(x+3)121+3x12+3x+x√x)
Step 6.2
Use n√ax=axn to rewrite √x as x12.
ln(5(x+3)121+3x12+3x+x⋅x12)
Step 6.3
Multiply x by x12 by adding the exponents.
Step 6.3.1
Multiply x by x12.
Step 6.3.1.1
Raise x to the power of 1.
ln(5(x+3)121+3x12+3x+x1x12)
Step 6.3.1.2
Use the power rule aman=am+n to combine exponents.
ln(5(x+3)121+3x12+3x+x1+12)
ln(5(x+3)121+3x12+3x+x1+12)
Step 6.3.2
Write 1 as a fraction with a common denominator.
ln(5(x+3)121+3x12+3x+x22+12)
Step 6.3.3
Combine the numerators over the common denominator.
ln(5(x+3)121+3x12+3x+x2+12)
Step 6.3.4
Add 2 and 1.
ln(5(x+3)121+3x12+3x+x32)
ln(5(x+3)121+3x12+3x+x32)
Step 6.4
Reorder terms.
ln(5(x+3)123x+3x12+x32+1)
Step 6.5
Rewrite 3x+3x12+x32+1 in a factored form.
Step 6.5.1
Factor 3x12 out of 3x+3x12.
Step 6.5.1.1
Factor 3x12 out of 3x.
ln(5(x+3)123x12(x12)+3x12+x32+1)
Step 6.5.1.2
Factor 3x12 out of 3x12.
ln(5(x+3)123x12(x12)+3x12(1)+x32+1)
Step 6.5.1.3
Factor 3x12 out of 3x12(x12)+3x12(1).
ln(5(x+3)123x12(x12+1)+x32+1)
ln(5(x+3)123x12(x12+1)+x32+1)
Step 6.5.2
Rewrite x32 as (x12)3.
ln(5(x+3)123x12(x12+1)+(x12)3+1)
Step 6.5.3
Rewrite 1 as 13.
ln(5(x+3)123x12(x12+1)+(x12)3+13)
Step 6.5.4
Since both terms are perfect cubes, factor using the sum of cubes formula, a3+b3=(a+b)(a2-ab+b2) where a=x12 and b=1.
ln(5(x+3)123x12(x12+1)+(x12+1)((x12)2-x12⋅1+12))
Step 6.5.5
Simplify.
Step 6.5.5.1
Multiply the exponents in (x12)2.
Step 6.5.5.1.1
Apply the power rule and multiply exponents, (am)n=amn.
ln(5(x+3)123x12(x12+1)+(x12+1)(x12⋅2-x12⋅1+12))
Step 6.5.5.1.2
Cancel the common factor of 2.
Step 6.5.5.1.2.1
Cancel the common factor.
ln(5(x+3)123x12(x12+1)+(x12+1)(x12⋅2-x12⋅1+12))
Step 6.5.5.1.2.2
Rewrite the expression.
ln(5(x+3)123x12(x12+1)+(x12+1)(x1-x12⋅1+12))
ln(5(x+3)123x12(x12+1)+(x12+1)(x1-x12⋅1+12))
ln(5(x+3)123x12(x12+1)+(x12+1)(x1-x12⋅1+12))
Step 6.5.5.2
Simplify.
ln(5(x+3)123x12(x12+1)+(x12+1)(x-x12⋅1+12))
Step 6.5.5.3
Multiply -1 by 1.
ln(5(x+3)123x12(x12+1)+(x12+1)(x-x12+12))
Step 6.5.5.4
One to any power is one.
ln(5(x+3)123x12(x12+1)+(x12+1)(x-x12+1))
ln(5(x+3)123x12(x12+1)+(x12+1)(x-x12+1))
Step 6.5.6
Factor x12+1 out of 3x12(x12+1)+(x12+1)(x-x12+1).
Step 6.5.6.1
Factor x12+1 out of 3x12(x12+1).
ln(5(x+3)12(x12+1)(3x12)+(x12+1)(x-x12+1))
Step 6.5.6.2
Factor x12+1 out of (x12+1)(3x12)+(x12+1)(x-x12+1).
ln(5(x+3)12(x12+1)(3x12+x-x12+1))
ln(5(x+3)12(x12+1)(3x12+x-x12+1))
Step 6.5.7
Subtract x12 from 3x12.
ln(5(x+3)12(x12+1)(x+2x12+1))
Step 6.5.8
Rewrite x as (x12)2.
ln(5(x+3)12(x12+1)((x12)2+2x12+1))
Step 6.5.9
Let u=x12. Substitute u for all occurrences of x12.
ln(5(x+3)12(x12+1)(u2+2u+1))
Step 6.5.10
Factor using the perfect square rule.
Step 6.5.10.1
Rewrite 1 as 12.
ln(5(x+3)12(x12+1)(u2+2u+12))
Step 6.5.10.2
Check that the middle term is two times the product of the numbers being squared in the first term and third term.
2u=2⋅u⋅1
Step 6.5.10.3
Rewrite the polynomial.
ln(5(x+3)12(x12+1)(u2+2⋅u⋅1+12))
Step 6.5.10.4
Factor using the perfect square trinomial rule a2+2ab+b2=(a+b)2, where a=u and b=1.
ln(5(x+3)12(x12+1)(u+1)2)
ln(5(x+3)12(x12+1)(u+1)2)
Step 6.5.11
Replace all occurrences of u with x12.
ln(5(x+3)12(x12+1)(x12+1)2)
ln(5(x+3)12(x12+1)(x12+1)2)
ln(5(x+3)12(x12+1)(x12+1)2)
Step 7
Step 7.1
Multiply x12+1 by (x12+1)2.
Step 7.1.1
Raise x12+1 to the power of 1.
ln(5(x+3)12(x12+1)1(x12+1)2)
Step 7.1.2
Use the power rule aman=am+n to combine exponents.
ln(5(x+3)12(x12+1)1+2)
ln(5(x+3)12(x12+1)1+2)
Step 7.2
Add 1 and 2.
ln(5(x+3)12(x12+1)3)
ln(5(x+3)12(x12+1)3)