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Solutions...
6/{(1+log x)(5-log x)}+1/(1+log x)=1
method of this equation..
6/{(1+log x)(5-log x)}+1/(1+log x)=1
method of this equation..
令1+logx=y(y不等于0）,则原式变为6/{y*(6-y)}+1/y=1；
等式通分后,变为6+(6-y)=y*(6-y)；
移项合并后,变为y^2-7y+12=0；
则y=3或者y=4,即logx=2或logx=3,查表可得x.
if 1+logx=y(y is not 0）,then the equation becomes 6/{y*(6-y)}+1/y=1；
then change into:6+(6-y)=y*(6-y)；
then:y^2-7y+12=0；
finally,we find y=3 or y=4,then logx=2 or logx=3,then check the logarithm table,you can find x.
等式通分后,变为6+(6-y)=y*(6-y)；
移项合并后,变为y^2-7y+12=0；
则y=3或者y=4,即logx=2或logx=3,查表可得x.
if 1+logx=y(y is not 0）,then the equation becomes 6/{y*(6-y)}+1/y=1；
then change into:6+(6-y)=y*(6-y)；
then:y^2-7y+12=0；
finally,we find y=3 or y=4,then logx=2 or logx=3,then check the logarithm table,you can find x.
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