Solving this logarithmic problemStuck on an 'advanced logarithm problem': $2 log_2 x - log_2 (x - tfrac1 2) =...
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Solving this logarithmic problem
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Solving this logarithmic problem
Stuck on an 'advanced logarithm problem': $2 log_2 x - log_2 (x - tfrac1 2) = log_3 3$Solve logarithmic simultaneous equations.Logarithmic EquationFind the roots of the quadratics functionLinear Equation Problem answerSimple Logarithmic question.Solving exponential equation: $-6+3e^x=8$Laws of logarithms: why isn't $frac{1}{4}log_2(8x - 56)^{16 }- 12 = log_2((8x-56)^{16})^{frac 1 4} - 12$?solving this linear equation by substitution?Factorization of combinatorics equation
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Given this equation: $$ln(x^2-1)-3=ln(x+1)$$
Evaluate x.
Applying the natural logarithmic rule
$$ln(x+1)=3$$
$$x+1=e^3$$
$$x=e^3-1$$
The answer was different from the book. Where did I when wrong?
algebra-precalculus
asked 8 hours ago
TriangleTriangle
1174
$endgroup$
add a comment |
$begingroup$
Given this equation: $$ln(x^2-1)-3=ln(x+1)$$
Evaluate x.
Applying the natural logarithmic rule
$$ln(x+1)=3$$
$$x+1=e^3$$
$$x=e^3-1$$
The answer was different from the book. Where did I when wrong?
algebra-precalculus
asked 8 hours ago
TriangleTriangle
1174
$endgroup$
$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago
add a comment |
$begingroup$
Given this equation: $$ln(x^2-1)-3=ln(x+1)$$
Evaluate x.
Applying the natural logarithmic rule
$$ln(x+1)=3$$
$$x+1=e^3$$
$$x=e^3-1$$
The answer was different from the book. Where did I when wrong?
algebra-precalculus
asked 8 hours ago
TriangleTriangle
1174
$endgroup$
Given this equation: $$ln(x^2-1)-3=ln(x+1)$$
Evaluate x.
Applying the natural logarithmic rule
$$ln(x+1)=3$$
$$x+1=e^3$$
$$x=e^3-1$$
The answer was different from the book. Where did I when wrong?
algebra-precalculus
algebra-precalculus
asked 8 hours ago
TriangleTriangle
1174
asked 8 hours ago
TriangleTriangle
1174
asked 8 hours ago
TriangleTriangle
1174
asked 8 hours ago
TriangleTriangle
1174
asked 8 hours ago
TriangleTriangle
1174
1174
$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago
add a comment |
$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago
$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago
$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
$endgroup$
add a comment |
$begingroup$
$ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$
implies $ln(x-1)=3$, $x=e^3+1$
edited 5 hours ago
Andrew Li
4,19621128
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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votes
$begingroup$
You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
$endgroup$
add a comment |
$begingroup$
You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
$endgroup$
add a comment |
$begingroup$
You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
$endgroup$
You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
answered 8 hours ago
Spencer KraislerSpencer Kraisler
666312
666312
add a comment |
add a comment |
$begingroup$
$ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$
implies $ln(x-1)=3$, $x=e^3+1$
edited 5 hours ago
Andrew Li
4,19621128
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
$endgroup$
add a comment |
$begingroup$
$ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$
implies $ln(x-1)=3$, $x=e^3+1$
edited 5 hours ago
Andrew Li
4,19621128
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
$endgroup$
add a comment |
$begingroup$
$ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$
implies $ln(x-1)=3$, $x=e^3+1$
edited 5 hours ago
Andrew Li
4,19621128
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
$endgroup$
$ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$
implies $ln(x-1)=3$, $x=e^3+1$
edited 5 hours ago
Andrew Li
4,19621128
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
edited 5 hours ago
Andrew Li
4,19621128
edited 5 hours ago
Andrew Li
4,19621128
edited 5 hours ago
Andrew Li
4,19621128
4,19621128
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
answered 8 hours ago
Tsemo AristideTsemo Aristide
63.1k11547
63.1k11547
add a comment |
add a comment |
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$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago