Solve logarithmic with Radical equation l Simple Step in This Log Problem: Can You Solve it
InfiniteIntegers и Inmathsexplorer
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Solve logarithmic with Radical equation l Simple Step in This Log Problem: Can You Solve it
1 121 просмотр · 3 дня назад
InfiniteIntegers и Inmathsexplorer
1 121 просмотр · 3 дня назад
Can you solve this logarithmic radical equation without getting trapped by the fraction 5/6?
Here is the challenge:
log_4(∛(2x²)) = 5/6
At first, this looks like a complicated combination of logarithms, radicals, fractional exponents, and algebra. But there is a hidden structure that makes the equation much easier than it appears.
The key is to recognize that 4 = 2².
First, convert the logarithmic equation into exponential form:
∛(2x²) = 4^(5/6)
Then rewrite the power of 4:
4^(5/6) = (2²)^(5/6) = 2^(5/3)
Now the cube root and fractional exponent fit together perfectly. Cubing both sides gives:
2x² = 2^5
So:
x² = 16
Therefore:
x = ±4
But the solution is not complete until we verify both values in the original equation.
In this video, we solve the equation step by step, expose the hidden trick, check the real-valued restrictions, and verify both solutions directly.
Challenge: Would you have noticed the connection between the cube root and the fraction 5/6 before starting the calculation?
Comment your answer below and tell me which step you noticed first.
Topics covered:
logarithmic equations, radical equations, solving for x, logarithm properties, exponential form, fractional exponents, cube roots, algebraic equations, real solutions, logarithm and radical hybrids.
If you enjoy mathematical problems that look simple but hide an elegant transformation, subscribe for more challenging logarithm and algebra problems.
#Logarithms #MathChallenge #SolveForX #Algebra #Mathematics
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