Can you solve this system of equations

Algebra Level 4

Given the following system of equations:- $a^{m}b^{n}=x$ $c^{m}d^{n}=y$ Where a,b,c,d,x,y,m and n are non zero positive real numbers Find out the value of m+n

\frac{log(\frac{x}{y})log(a)+log(\frac{y}{x})log(b)}{log(x)-log(y)}

\frac{log(a)log(c)-log(b)log(d)}{log(x)-log(y)}

\frac{log(x)log(\frac{d}{c})-log(y)log(\frac{b}{a})}{log(a)log(d)-log(b)log(c)}

\frac{log(ab)-log(bc)}{log(x)log(a)-log(y)log(b)}

1 solution

Mateus Gomes
Feb 5, 2016

$\Rightarrow {\color{#D61F06}{log(a^mb^n)=log(x)}}$ $\Rightarrow mlog(a)+nlog(b)=log(x)$ $\Rightarrow n=\frac{log(x)-mlog(a)}{log(b)}$ $\Rightarrow m=\frac{log(x)-nlog(b)}{log(a)}$ $\Rightarrow {\color{#3D99F6}{log(c^md^n)=log(y)}}$ $\Rightarrow mlog(c)+nlog(d)=log(y)$ $\Rightarrow mlog(c)+\frac{log(x)-mlog(a)}{log(b)}log(d)=log(y)$ $\Rightarrow mlog(c)log(b)+[log(x)-mlog(a)] log(d)=log(y)log(b)$ $\Rightarrow {\color{#302B94}{m=\frac{log(y)log(b)-log(d)log(x)}{log(c)log(b)-log(d)log(a)}}}$ $\Rightarrow \frac{log(x)-nlog(b)}{log(a)}log(c)+nlog(d)=log(y)$ $\Rightarrow [log(x)-nlog(b)]log(c)+ nlog(d)log(a)=log(y)log(a)$ $\Rightarrow {\color{#20A900}{n=\frac{log(x)log(c)-log(y)log(a)}{log(c)log(b)-log(d)log(a)}}}$ $\Rightarrow {\color{#EC7300}{m+n=\frac{log(y)log(b)-log(d)log(x)+log(x)log(c)-log(y)log(a)}{log(c)log(b)-log(d)log(a)}}}$ $\Rightarrow {\color{#3D99F6}{\boxed{m+n=\frac{log(x)log(\frac{d}{c})-log(y)log(\frac{b}{a})}{log(a)log(d)-log(b)log(c)}}}}$

Can you solve this system of equations

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