More logarithm problems

  • Thread starter mathdummy
  • Start date
  • #1
[tex]log_10x-2[/tex]=0 ... that's log base 10
(answer: 100)

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

[tex]log_4x-log_4(x-1)[/tex]=1/2
(answer: 2)
 
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Answers and Replies

  • #2
futz
80
0
[tex]
\ln(x+5)=\ln(x-1)-\ln(x+1)=\ln\left(\frac{x-1}{x+1}\right)
[/tex]

So,

[tex]
x+5=\frac{x-1}{x+1}
[/tex]

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
PrudensOptimus
641
0
#1:


[tex]\log_{10} x - 2 = 0[/tex]


[tex]10^{\log_{10}x} = 10^{2}[/tex]


[tex]x = 10^{2} = 100[/tex]

#3:


[tex]\log_{4}x - \log_{4}(x-1) = 0.5[/tex]

[tex]\log_{4}\frac{x}{(x-1)} = 0.5[/tex]

[tex]\frac{x}{(x-1)} = 2[/tex]

[tex]x = 2x - 2[/tex]

[tex]x = 2 (x > 1)[/tex]
 
Last edited:

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