# Thread: Relationship between a^x and log to the base a, integration.

1. ## Relationship between a^x and log to the base a, integration.

Describe the relationship between $\displaystyle a^x$ and log to the base a, giving the exponential function as an example. Simplify the function
$\displaystyle \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}$

and hence evaluate the integral

$\displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx$

2. Originally Posted by Boz

Describe the relationship between $\displaystyle a^x$ and log to the base a, giving the exponential function as an example. Simplify the function
$\displaystyle \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}$

and hence evaluate the integral

$\displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx$

note that ...

$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$

$\displaystyle \ln(e^{3x}) = 3x$ i.e. ... $\displaystyle \log_a(a^x) = x$

you will end up with the definite integral $\displaystyle \displaystyle \int_0^1 3x \cdot e^x \, dx$ which can be evaluated using integration by parts.

3. Originally Posted by Boz

Describe the relationship between $\displaystyle \displaystyle a^x$ and log to the base a, giving the exponential function as an example. Simplify the function
$\displaystyle \displaystyle \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}$

and hence evaluate the integral

$\displaystyle \displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx$
Is that not easier to read?

4. Originally Posted by Boz

Describe the relationship between $\displaystyle a^x$ and log to the base a, giving the exponential function as an example.
$\displaystyle log_a(x)$ and $\displaystyle a^x$ are inverse functions: $\displaystyle log_a(a^x)= x$ and $\displaystyle a^{log_a(x)}= x$

Simplify the function
$\displaystyle \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}$
$\displaystyle ln(e^3x)= 3x$, $\displaystyle (e^{cos x})^{2sin x}= e^{2sin x cos x}= e^{sin 2x}$ so $\displaystyle \frac{e^{sin 2x}lne^{3x}}{(e^{cosx})^{2sinx}e^{-x}}= \frac{e^{sin 2x}3x}{e^{sin 2x}}e^{-x}= 3xe^{-x}$

Integrate using integration by parts: $\displaystyle u= e^{-x}$, $\displaystyle dv= 3x$.

and hence evaluate the integral

$\displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx$

5. Originally Posted by hallsofivy
$\displaystyle log_a(x)$ and $\displaystyle a^x$ are inverse functions: $\displaystyle log_a(a^x)= x$ and $\displaystyle a^{log_a(x)}= x$

$\displaystyle ln(e^3x)= 3x$, $\displaystyle (e^{cos x})^{2sin x}= e^{2sin x cos x}= e^{sin 2x}$ so $\displaystyle \frac{e^{sin 2x}lne^{3x}}{(e^{cosx})^{2sinx}e^{-x}}= \frac{e^{sin 2x}3x}{e^{sin 2x}}e^{-x}= 3xe^{-x}$

integrate using integration by parts: $\displaystyle u= e^{-x}$, $\displaystyle dv= 3x$.

And hence evaluate the integral

$\displaystyle \int^1_0 \frac{e^{sin2x} lne^{3x}}{{(e^{cosx})}^{2sinx}e^{-x}}dx$

$\displaystyle \displaystyle \int 3xe^x dx$
Let $\displaystyle u=3x, \frac{du}{dx}=3,$$\displaystyle du=3dx, Let ,dv=e^x dx, v=\int e^x dx=e^x$
$\displaystyle \displaystyle 3xe^x-\int e^x 3dx, 3xe^x-3e^x=3e^x(x-1)$
Hence $\displaystyle \displaystyle \int_0^1 3xe^x dx= [3e^x(x-1)] 0 to 1= [3e^1(1-1)]-[3e^0(0-1)]=-3*(-1)=3$