# Logarithmic Proof (23/03/2015)

Prove that:

$\ln { \left( xy \right) } =\ln { x+\ln { y } }$

Now, say that:

${ a }^{ m }=x\\ \\ \therefore \quad \log _{ a }{ x=m } \\ \\ { a }^{ q }=y\\ \\ \therefore \quad \log _{ a }{ y } =q$

So…

$LHS\\ \\ =\ln { \left( { a }^{ m }\cdot { a }^{ q } \right) } \\ \\ =\ln { \left( { a }^{ \left( m+q \right) } \right) } \\ \\ =\left( m+q \right) \cdot \ln { a } \\ \\ =\left( \log _{ a }{ x } +\log _{ a }{ y } \right) \cdot \log _{ e }{ a } \\ \\ =\frac { \log _{ a }{ x } +\log _{ a }{ y } }{ \log _{ a }{ e } } \\ \\ =\frac { \log _{ a }{ x } }{ \log _{ a }{ e } } +\frac { \log _{ a }{ y } }{ \log _{ a }{ e } } \\ \\ =\log _{ e }{ x } +\log _{ e }{ y } \\ \\ =\ln { x } +\ln { y } \\ \\ =RHS$

# New Logarithmic Proof – Quicker Version

Prove that:

$\log _{ a }{ x } =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } }$

————-

PROOF:

$RHS\\ \\ =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } } \\ \\ =\log _{ b }{ x } \cdot \log _{ a }{ b } \\ \\ =\log _{ a }{ \left( { b }^{ \log _{ b }{ x } } \right) } \\ \\ =\log _{ a }{ x } \\ \\ =LHS$

# Logarithmic Differentiation As Seen In Video…

In this video, Patrick JMT demonstrated how to do logarithmic differentiation.

Now I’m going to show you how I’d solve the same problem…

Firstly we must know that:

$y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ If\quad y=u\cdot v,\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ y=x\ln { \left( \ln { x } \right) } =u\cdot v\\ \\ u=x,\quad \frac { du }{ dx } =1\\ \\ v=\ln { \left( \ln { x } \right) } =\ln { q } \\ \\ \frac { dv }{ dq } =\frac { 1 }{ q } =\frac { 1 }{ \ln { x } } \\ \\ q=\ln { x } ,\quad \frac { dq }{ dx } =\frac { 1 }{ x } \\ \\ \therefore \quad \frac { dv }{ dx } =\frac { 1 }{ x\ln { x } } \\ \\ \therefore \quad \frac { dy }{ dx } =x\cdot \frac { 1 }{ x\ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) }$

Now:

$y={ \left( \ln { x } \right) }^{ x }\\ \\ \ln { y=\ln { \left( { \left( \ln { x } \right) }^{ x } \right) } } \\ \\ \ln { y } =x\ln { \left( \ln { x } \right) } \\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\} \\ \\ \frac { dy }{ dx } ={ \left( \ln { x } \right) }^{ x }\left\{ \frac { 1 }{ \ln { x } } +\ln { \left( \ln { x } \right) } \right\}$

# If y=ln(f(x)), dy/dx=(f'(x))/f(x)

Prove that if:

$y=\ln { \left( f\left( x \right) \right) } ,\quad \frac { dy }{ dx } =\frac { f$

Firstly, say that:

$y=h\left( x \right) \\ \\ g=f\left( x \right)$

So:

# Tough Logarithmic Problem…

Prove that:

$\log _{ a }{ x } =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } }$

Say that:

$\log _{ b }{ x } =p\\ \\ \therefore \quad { b }^{ p }=x$

And that:

$\log _{ b }{ a } =q\\ \\ \therefore \quad { b }^{ q }=a$

Therefore:

$\log _{ a }{ \left( x \right) } \\ \\ =\log _{ a }{ \left( { b }^{ p } \right) } \\ \\ =p\log _{ a }{ \left( b \right) } \\ \\ =p\cdot \frac { 1 }{ \log _{ b }{ a } } \\ \\ =\frac { \log _{ b }{ x } }{ \log _{ b }{ a } }$

# If y=a^x, dy/dx=a^xlna proof

${ a }^{ x }=y\\ \\ \log _{ a }{ y=x } \\ \\ \frac { \ln { y } }{ \ln { a } } =x\\ \\ x=\frac { 1 }{ \ln { a } } \cdot \ln { y } \\ \\ x=n\cdot u\\ \\ \therefore \quad \frac { dx }{ du } =n\\ \\ u=\ln { y,\quad \frac { du }{ dy } } =\frac { 1 }{ y } \\ \\ \frac { dx }{ dy } =\frac { dx }{ du } \cdot \frac { du }{ dy } =\frac { n }{ y } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \frac { dx }{ dy } } \\ \frac { dy }{ dx } =\frac { 1 }{ \frac { n }{ y } } \\ \frac { dy }{ dx } =\frac { y }{ n } \\ \\ but,\quad y={ a }^{ x }\quad and\quad n=\frac { 1 }{ \ln { a } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { { a }^{ x } }{ \frac { 1 }{ \ln { a } } } \\ \\ \frac { dy }{ dx } ={ a }^{ x }\cdot \ln { a } \\ \\ \\$

# If y=lnx, dy/dx=1/x proof

$\log _{ e }{ x } =y\\ \\ \therefore \quad x={ e }^{ y }\\ \\ \frac { dx }{ dy } ={ e }^{ y }\\ \\$

Now:

$\frac { dy }{ dx } =\frac { 1 }{ \frac { dx }{ dy } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ { e }^{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ x } \\ \\$

# Logarithmic Proof (4)

Prove that:

$\log _{ a }{ \left( \frac { x }{ p } \right) } =\log _{ a }{ \left( x \right) } -\log _{ a }{ \left( p \right) }$

Say that:

$\log _{ a }{ \left( x \right) } =m\\ \\ \therefore \quad { a }^{ m }=x$

And that:

$\log _{ a }{ \left( p \right) } =n\\ \\ \therefore \quad { a }^{ n }=p$

Therefore:

$\log _{ a }{ \left( \frac { x }{ p } \right) } \\ \\ =\log _{ a }{ \left( \frac { { a }^{ m } }{ { a }^{ n } } \right) } \\ \\ =\log _{ a }{ \left( { a }^{ \left( m-n \right) } \right) } \\ \\ =\left( m-n \right) \log _{ a }{ \left( a \right) } \\ \\ =m\log _{ a }{ \left( a \right) } -n\log _{ a }{ \left( a \right) } \\ \\ =\log _{ a }{ \left( { a }^{ m } \right) -\log _{ a }{ \left( { a }^{ n } \right) } } \\ \\ =\log _{ a }{ \left( x \right) } -\log _{ a }{ \left( p \right) }$

# Logarithmic Proof (3)

Prove that:

$\log _{ a }{ \left( { a }^{ n } \right) =n\log _{ a }{ (a) } }$

Say that:

${ a }^{ n }=p$

Therefore:

$\log _{ a }{ p=n }$

So:

$\log _{ a }{ \left( { a }^{ n } \right) } \\ \\ =\log _{ a }{ \left( p \right) } \\ \\ =\log _{ a }{ \left( p \right) } \cdot 1\\ \\ =\log _{ a }{ \left( p \right) \cdot \log _{ a }{ \left( a \right) } } \\ \\ =n\cdot \log _{ a }{ \left( a \right) }$