SOLUTION: Solve the equation: loga^x-loga^(x-5)=loga^6
log of a to x minus log of a to x-5 equals lof of a to the six
We are supposed to use the properties of logarithms to solve this prob
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Exponential-and-logarithmic-functions
-> SOLUTION: Solve the equation: loga^x-loga^(x-5)=loga^6
log of a to x minus log of a to x-5 equals lof of a to the six
We are supposed to use the properties of logarithms to solve this prob
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Question 587847: Solve the equation: loga^x-loga^(x-5)=loga^6
log of a to x minus log of a to x-5 equals lof of a to the six
We are supposed to use the properties of logarithms to solve this problem. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Solve the equation: loga^x-loga^(x-5)=loga^6
log of a to x minus log of a to x-5 equals lof of a to the six
We are supposed to use the properties of logarithms to solve this problem.
**
loga^x-loga^(x-5)=loga^6
loga^x-loga^(x-5)-loga^6=0
loga^x-(loga^(x-5)+loga^6)=0
place under a single log
log[(a^x)/a^(x-5)*a^6]=0
convert to exponential form: base(10) raised to log of number(0)=number(a^x)/a^(x-5)*a^6
10^0=(a^x)/a^(x-5)*a^6=1
(a^x)=a^(x-5)*a^6
a^x=a^(x-5+6)=a^(x+1)
x=x+1
0≠1 (no solution)
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