# To find value of alogb−logcblogc−logacloga−logba^{\log b-\log c } b^{\log c-\log a } c^{\log a -\log b}

To find value of

alogb−logcblogc−logacloga−logba^{\log b -\log c } b^{\log c -\log a} c^{\log a -\log b } where a,b,ca, b, c are positive real numbers. Now using formulas

i write this as (bc)loga(ca)logb(ab)logc(\frac{b}{c})^{\log a }(\frac{c}{a})^{\log b }(\frac{a}{b})^{\log c }

How do i proceed further from here?

Thanks

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Downvoter explain your hatred !
– J. Deff
2 days ago

Try using xlogy=ylogxx^{\log y} = y^{\log x} after applying xy−z=xy/xzx^{y – z} = x^y / x^z
– DanielV
2 days ago

This is what i have done
– J. Deff
2 days ago

If you do it to the correct terms, they will cancel out. You will have equal terms in the numerator and denominator.
– DanielV
2 days ago

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I assume that your formula is alnb−lncblnc−lnaclna−lnba^{\ln b-\ln c} b^{\ln c-\ln a} c^{\ln a-\ln b}. Let x=x= this number, take ln\ln on the both of sides, we get lnx=(lnb−lnc)lna+(lnc−lna)lnb+(lna−lnb)lnc=0\ln x=(\ln b-\ln c)\ln a+(\ln c-\ln a)\ln b+(\ln a-\ln b)\ln c=0, then x=1x=1.

How to do this without taking log on both sides
– J. Deff
2 days ago

use ax=exlnaa^x=e^{x\ln a}
– Mingfeng Zhao
2 days ago