Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$${ a }^{ x } =bc, { b }^{ y } =ca, { c }^{ z } =ab, \frac{ x }{ 1+x } + \frac{ y }{ 1+y } + \frac{ z }{ 1+z } =2$$

Answer

$$x=log(a,b*c),log(a,b^y)$$

Solution


Rewrite the expression with a common denominator.
\[\begin{aligned}&{a}^{x}=bc\\&{b}^{y}=ca\\&{c}^{z}=ab\\&\frac{x(1+y)(1+z)+y(1+x)(1+z)+z(1+x)(1+y)}{(1+x)(1+y)(1+z)}=2\end{aligned}\]
Expand.
\[\begin{aligned}&{a}^{x}=bc\\&{b}^{y}=ca\\&{c}^{z}=ab\\&\frac{x+xz+xy+xyz+y+yz+yx+yxz+z+zy+zx+zxy}{(1+x)(1+y)(1+z)}=2\end{aligned}\]
Collect like terms.
\[\begin{aligned}&{a}^{x}=bc\\&{b}^{y}=ca\\&{c}^{z}=ab\\&\frac{x+(xz+xz)+(xy+xy)+(xyz+xyz+xyz)+y+(yz+yz)+z}{(1+x)(1+y)(1+z)}=2\end{aligned}\]
Simplify  \(x+(xz+xz)+(xy+xy)+(xyz+xyz+xyz)+y+(yz+yz)+z\)  to  \(x+2xz+2xy+3xyz+y+2yz+z\).
\[\begin{aligned}&{a}^{x}=bc\\&{b}^{y}=ca\\&{c}^{z}=ab\\&\frac{x+2xz+2xy+3xyz+y+2yz+z}{(1+x)(1+y)(1+z)}=2\end{aligned}\]
Break down the problem into these 2 equations.
\[{a}^{x}=bc\]
\[{a}^{x}={b}^{y}\]
Solve the 1st equation: \({a}^{x}=bc\).
\[x=\log_{a}{(bc)}\]
Solve the 2nd equation: \({a}^{x}={b}^{y}\).
\[x=\log_{a}{{b}^{y}}\]
Collect all solutions.
\[x=\log_{a}{(bc)},\log_{a}{{b}^{y}}\]
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