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Solve the equation for x : log to the base a ( 1 + log to the base b ( 1 + log to the base c ( 1 + log to the base d (x)))) = 0
take antilog on both sides,which gives... 1 + log to the base b ( 1 + log to the base c ( 1 + log to the base d (x)))= a0. log to the base b ( 1 + log to the base c ( 1 + log to the base d (x))) = 0....take antilog again. 1 + log to the base c ( 1 + log to the base d (x)) = b0.log to the base c ( 1 + log to the base d (x)) = 0....take antilog again.1 + log to the base d (x) = c0. log to the base d (x) = 0....take antilog again.x=d0.x=1....ANS.REMEMBER...logax=0...on taking antilog it converts to.x=a0....which gives x=1
loga(1+logb(1+logc(1+logd (x)=0take antilog on both sides..1+logb(1+logc(1+logd (x)=a0logb(1+logc(1+logd (x)=0take antilog on both sides1+logc(1+logd (x)=b0logc(1+logd (x)=0.take antilog on both sides1+logd (x)=c0logd (x)=0take antilog x=d0x=1..ANS[REMEMBER.........loga b=0...on taking antilogb=a to the power of 0b=1]
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