log(x-1)+log(x+1)=log2 1
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Answer:
There are no solutions.
Explanation:
Use the logarithm rules to simplify either side:
Left hand side: loga+logb=log(ab)Right hand side: bloga=log(ab)
This gives
log[(x−1)(x+1)]=log[(x+2)2]
This can be simplified using the following rule:
If loga=logb, then a=b
Giving us:
(x−1)(x+1)=(x+2)2
Distribute both of these.
x2−1=x2+4x+4
Solve. The x2 terms will cancel, so there will only be one solution.
4x=−5
x=−54
However, this solution is invalid. Imagine if xactually were −54. Plug it into the original equation. The terms log(x−1) and log(x+1)would be log(−94) and log(−14), and the logarithm function loga is only defined when a>0. I THINK THIS WILL HELP YOU ! THANK YOU ☺☺
There are no solutions.
Explanation:
Use the logarithm rules to simplify either side:
Left hand side: loga+logb=log(ab)Right hand side: bloga=log(ab)
This gives
log[(x−1)(x+1)]=log[(x+2)2]
This can be simplified using the following rule:
If loga=logb, then a=b
Giving us:
(x−1)(x+1)=(x+2)2
Distribute both of these.
x2−1=x2+4x+4
Solve. The x2 terms will cancel, so there will only be one solution.
4x=−5
x=−54
However, this solution is invalid. Imagine if xactually were −54. Plug it into the original equation. The terms log(x−1) and log(x+1)would be log(−94) and log(−14), and the logarithm function loga is only defined when a>0. I THINK THIS WILL HELP YOU ! THANK YOU ☺☺
honeybee05:
hey ur answee is wrong
then what is the answer?
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