How do you solve #log x - log(x-10)=1#?

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Answer 1 · 2015-08-10T00:56:47

#color(red)(x=100/9)#

# logx-log(x−10)=1#

Recall that #loga-logb=log(a/b)#, so

# logx-log(x−10)=log(x/(x-10))#

# log(x/(x-10))=1#

Convert the logarithmic equation to an exponential equation.

#10^(log(x/(x-10))) = 10^1#

Remember that #10^logx =x#, so

#x/(x-10)=10#

#x=10(x-10)#

#x=10x-100#

#9x=100#

#x=100/9#

Check:

# logx-log(x−10)=1#

If #x=100/9#,

# log(100/9)-log(100/9−10)=1#

#log(100/9)-log(100/9-90/9)=1#

#log(100/9)-log((100-90)/9)=1#

#log(100/9)-log(10/9)=1#

#log((100/color(red)(cancel(color(black)(9))))/(10/color(red)(cancel(color(black)(9)))))=1#

#log(100/10)=1#

#log10 = 1#

#1=1#

#x=100/9# is a solution.