User:Raymonm

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Proofs of Every Logarithm Property

$1$ : $\text{log}_a b^n=n~\text{log}_a b$

Proof: Let $x=\text{log}_a b^n$ and $y=n~\text{log}_a b$ . This implies that $a^x=b^n$ and $\frac{y}{n}=\text{log}_a b\implies a^{\frac{y}{n}}=b$ . Raising the equation $a^{\frac{y}{n}}=b$ to the $n^\text{th}$ power on both sides gives $a^y=b^n$ . If $a^x=b^n$ and $a^y=b^n$ , then $x=y$ . This means that $\text{log}_a b^n=n~\text{log}_a b$ .

$2$ : $\text{log}_a b+\text{log}_a c=\text{log}_a bc$

Proof: If $x=\text{log}_a b$ and $y=\text{log}_a c$ and $z=\text{log}_a bc$ , then $x+y=z$ . By the definition of a logarithm, $a^x=b$ , $a^y=c$ , and $a^z=bc$ . Multiplying the first two equations together, $a^x a^y=bc\implies a^{x+y}=bc$ . Since $a^{x+y}=bc$ and $a^z=bc$ , then $x+y=z$ .

$3$ : $\text{log}_a b-\text{log}_a c=\text{log}_a \frac{b}{c}$

Proof: If $x=\text{log}_a b$ and $y=\text{log}_a c$ and $z=\text{log}_a \frac{b}{c}$ , then $x-y=z$ . Then, $a^x=b$ , $a^y=c$ , and $a^z=\frac{b}{c}$ . We divide the first two equations together to get $\frac{a^x}{a^y}=\frac{b}{c}\implies a^{x-y}=\frac{b}{c}$ . Because that $a^{x-y}=\frac{b}{c}$ and $a^z=\frac{b}{c}$ , $x-y=z$ .

$4$ : $(\text{log}_a b)(\text{log}_c d)=(\text{log}_a d)(\text{log}_c b)$

Proof: Let $\text{log}_a b=w$ , $\text{log}_c d=x$ , $\text{log}_a d=y$ , and $\text{log}_c b=z$ . We have to prove that $wx=yz$ . We write the logarithms in exponential form: $a^w=b$ , $c^x=d$ , $a^y=d$ , and $c^z=b$ . Combining the equations together in pairs by multiplication, $a^w c^x=a^y c^z=bd$ . This is true if and only if $w=y$ and $x=z$ . Combining the two equations together by multiplication gives $wx=yz$ .

$5$ : $\frac{\text{log}_a b}{\text{log}_a c}=\text{log}_c b$

Proof: We multiply both sides by $\text{log}_a c$ to get $\text{log}_a b=(\text{log}_a c)(\text{log}_c b)$ . If $\text{log}_a b=x$ , $\text{log}_a c=y$ , and $\text{log}_c b=z$ , $x=yz$ . Furthermore, $a^x=b$ , $a^y=c$ , and $c^z=b$ . Since $a^x=b$ and $c^z=b$ , $a^x=c^z$ . Raising both sides of the equation $a^y=c$ to the $z^\text{th}$ power gives $(a^y)^z=c^z\implies a^{yz}=c^z$ . Since $a^x=c^z$ and $a^{yz}=c^z$ , $a^x=a^{yz}$ . This is only true if $x=yz$ , so we are done.

$6$ : $\text{log}_{a^n} b^n=\text{log}_a b$

Proof: Let $x=\text{log}_{a^n} b^n$ and $y=\text{log}_a b$ . This implies that $x=y$ . Converting the logarithms into exponential form gives $(a^n)^x=b^n\implies a^{xn}=b^n \implies (a^x)^n=b^n$ and $a^y=b$ . Taking the $n^\text{th}$ root on both sides, $a^x=b$ . Since $a^x=a^y=b$ , $x=y$ .

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User:Raymonm