What is 3e^{0.6x}+9ln(2x^5)+8cos(2x)-9 ?

The solution to 3e^{0.6x}+9ln(2x^5)+8cos(2x)-9 is 3e^{0.6x}+9ln(2)+45ln(x)+8cos(2x)-9

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simplify 3e^{0.6x}+9ln(2x^5)+8cos(2x)-9

simplify

3 e^{0.6 x} + 9 ln (2 x^{5}) + 8 cos (2 x) - 9

3 e^{0.6 x} + 9 ln (2) + 45 ln (x) + 8 cos (2 x) - 9

Solution steps

3 e^{0.6 x} + 9 ln (2 x^{5}) + 8 cos (2 x) - 9

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

9 ln (2 x^{5}) = 9 ln (2) + 9 ln (x^{5})

= 3 e^{0.6 x} + 9 ln (2) + 9 ln (x^{5}) + 8 cos (2 x) - 9

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (x^{5}) = 5 ln (x)

= 3 e^{0.6 x} + 9 ln (2) + 9 \cdot 5 ln (x) + 8 cos (2 x) - 9

Multiply the numbers:

9 \cdot 5 = 45

= 3 e^{0.6 x} + 9 ln (2) + 45 ln (x) + 8 cos (2 x) - 9

\frac{1}{2} ln (x - 1) + \frac{1}{2} ln (x + 1)

- 4 lo g_{10} (x) - 2 lo g_{10} (y) + 100

\frac{1}{2} (ln (3) - 3 ln (2) + 3 ln (1))

lo g_{3} (\frac{y}{9})

lo g_{10} ((u v^{5})^{3})

What is 3e^{0.6x}+9ln(2x^5)+8cos(2x)-9 ?
The solution to 3e^{0.6x}+9ln(2x^5)+8cos(2x)-9 is 3e^{0.6x}+9ln(2)+45ln(x)+8cos(2x)-9