- #31
nae99
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Pranav-Arora said:
OMG, such simple mistakes huh, ok
x= 6/2 = 3
x= -(-2) - 2/ 2 = 0/2 x= 0
2log(x-1) + logx = logx + log4
- Thread starter nae99
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SUMMARY
The discussion revolves around solving the logarithmic equation 2log(x-1) + logx = logx + log4. Participants emphasize the importance of canceling identical log terms on both sides, leading to the simplified equation log(x-1)^2 = log4. The solution involves applying logarithmic identities, resulting in the quadratic equation (x-1)^2 = 4, which yields the final solutions x = 3 and x = -1, with a reminder that x must be greater than 1 for the logarithm to be defined.
PREREQUISITES- Understanding of logarithmic identities, specifically log(a) + log(b) = log(ab) and log(a) - log(b) = log(a/b).
- Familiarity with algebraic manipulation, including solving quadratic equations.
- Knowledge of the properties of logarithms, particularly the definition of logarithmic functions.
- Basic skills in handling equations involving square roots.
- Study logarithmic identities in depth, focusing on their applications in solving equations.
- Practice solving quadratic equations using the quadratic formula and factoring methods.
- Explore the implications of logarithmic functions in real-world applications, such as in finance and science.
- Review the concept of domain restrictions in logarithmic functions to understand valid input values.
Students studying algebra, particularly those tackling logarithmic equations, as well as educators looking for examples of logarithmic problem-solving techniques.
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