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

Question

Christian Antunez · Accepted Answer

#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.

Sadie Ackerman · Answer

undefined To solve the equation $ \log(x) - \log(x - 10) = 1 $, you can use the properties of logarithms.

1. Start by applying the property $ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) $:

$$ \log(x) - \log(x - 10) = \log\left(\frac{x}{x - 10}\right) $$

2. Set the equation equal to 1:

$$ \log\left(\frac{x}{x - 10}\right) = 1 $$

3. Convert the logarithmic equation into its exponential form:

$$ 10^1 = \frac{x}{x - 10} $$

4. Solve for $ x $:

$$ 10 = \frac{x}{x - 10} $$

$$ 10(x - 10) = x $$

$$ 10x - 100 = x $$

$$ 10x - x = 100 $$

$$ 9x = 100 $$

$$ x = \frac{100}{9} $$

So, the solution to the equation $ \log(x) - \log(x - 10) = 1 $ is $ x = \frac{100}{9} $.