# More logarithm problems

1. Nov 22, 2003

### mathdummy

$$log_10x-2$$=0 ....... that's log base 10
(answer: 100)

ln(x+5)=ln(x-1)-ln(x+1)
(answer: no solution. but why?)

$$log_4x-log_4(x-1)$$=1/2
(answer: 2)

Last edited by a moderator: Nov 22, 2003
2. Nov 22, 2003

### futz

$$\ln(x+5)=\ln(x-1)-\ln(x+1)=\ln\left(\frac{x-1}{x+1}\right)$$

So,

$$x+5=\frac{x-1}{x+1}$$

Solve this for x, and you will see that both possible values generate a negative argument in the second logarithm, so there is no valid solution.

3. Nov 22, 2003

### PrudensOptimus

#1:

$$\log_{10} x - 2 = 0$$

$$10^{\log_{10}x} = 10^{2}$$

$$x = 10^{2} = 100$$

#3:

$$\log_{4}x - \log_{4}(x-1) = 0.5$$

$$\log_{4}\frac{x}{(x-1)} = 0.5$$

$$\frac{x}{(x-1)} = 2$$

$$x = 2x - 2$$

$$x = 2 (x > 1)$$

Last edited: Nov 22, 2003
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook