Question

if log x/b-c = log y/c-a = log z/a-b, prove that xyz = x^a y^b z^c = 1

Answer 1

$\large\underline{\sf{Solution-}}$

Given Logarithmic function is

$\rm :\longmapsto\:\dfrac{logx}{b - c} = \dfrac{logy}{c - a} = \dfrac{logz}{a - b}$

Let assume that

$\rm :\longmapsto\:\dfrac{logx}{b - c} = \dfrac{logy}{c - a} = \dfrac{logz}{a - b} = k$

So,

$\rm :\longmapsto\:\dfrac{logx}{b - c} = k$

$\rm :\longmapsto\:logx = k(b - c)$

$\rm :\longmapsto\:alogx = ak(b - c)$

$\bf\implies \:log {x}^{a} = k(ab - ac) - - - - (1)$

Also,

$\rm :\longmapsto\:\dfrac{logy}{c - a} = k$

$\rm :\longmapsto\:logy = k(c - a)$

$\rm :\longmapsto\:blogy = bk(c - a)$

$\bf\implies \:log {y}^{b} = k(cb - ba) - - - - (2)$

Also,

$\rm :\longmapsto\:\dfrac{logz}{a - b} = k$

$\rm :\longmapsto\:logz = k(a - b)$

$\rm :\longmapsto\:clogz = ck(a - b)$

$\bf\implies \:log {z}^{c} = k(ca - bc) - - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm :\longmapsto\:log {x}^{a} + log {y}^{b} + log {z}^{c} = k(ab - ac + bc - ab + ac - bc)$

$\rm :\longmapsto\:log {x}^{a} + log {y}^{b} + log {z}^{c} = k(0)$

$\rm :\longmapsto\:log {x}^{a} + log {y}^{b} + log {z}^{c} = 0$

$\rm :\longmapsto\:log[{x}^{a} \: {y}^{b} \: {z}^{c}] = 0$

$\rm :\longmapsto\:log[{x}^{a} \: {y}^{b} \: {z}^{c}] = log1$

$\rm\implies \:\boxed{\tt{ {x}^{a} \: {y}^{b} \: {z}^{c} =1 \: }}$

[ See from attachment xyz = 1 ]

We concluded that

$\rm\implies \:\boxed{\tt{ xyz\: =\: {x}^{a} \: {y}^{b} \: {z}^{c} =1 \: }}$

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FORMULA USED

$\boxed{\tt{ bloga \: = \: log {a}^{b}}}$

$\boxed{\tt{ loga + logb = logab \: }}$

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MORE TO KNOW

$\boxed{\tt{ log_{x}(x) = 1 \: }}$

$\boxed{\tt{ log_{x}( {x}^{y} ) = y \: }}$

$\boxed{\tt{ log_{ {x}^{z} }( {x}^{y} ) = \frac{y}{z} \: }}$

$\boxed{\tt{ log_{ {x}^{z} }( {w}^{y} ) = \frac{y}{z} log_{x}(w) \: }}$

$\boxed{\tt{ {a}^{ log_{a}(x) } = x}}$

$\boxed{\tt{ {a}^{ ylog_{a}(x) } = {x}^{y} }}$

$\boxed{\tt{ {e}^{ ylog_{e}(x) } = {x}^{y} }}$