\log_2(8x) = 5 - RoadRUNNER Motorcycle Touring & Travel Magazine
Solving log₂(8x) = 5: A Step-by-Step Guide with Full Explanation
Solving log₂(8x) = 5: A Step-by-Step Guide with Full Explanation
Understanding logarithms is essential in algebra and many areas of science and engineering. One commonly encountered equation is log₂(8x) = 5. Whether you're a student tackling logarithmic expressions or a teacher explaining key concepts, knowing how to solve such equations is invaluable.
In this article, we break down how to solve log₂(8x) = 5, explain the underlying principles, and highlight why this type of problem matters.
Understanding the Context
What is log₂(8x) = 5?
The expression log₂(8x) = 5 means: “To what power must 2 raise to get the value of 8x?”
This logarithmic equation translates into a simple exponential form, which makes it straightforward to solve.
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Key Insights
Step-by-Step Solution
Step 1: Convert to exponential form
Recall that the equation log_b(A) = C is equivalent to b^C = A.
Here, base b = 2, A = 8x, and C = 5. Therefore:
$$
2^5 = 8x
$$
Step 2: Evaluate 2⁵
Calculate the left side:
$$
32 = 8x
$$
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Step 3: Solve for x
Divide both sides by 8 to isolate x:
$$
x = rac{32}{8} = 4
$$
Final Answer:
$$
oxed{x = 4}
$$
Why This Equation Matters
- Logarithmic Equations Develop Log Skills: Mastering such equations strengthens your ability to work with variables inside logarithms, a crucial skill in advanced math and calculus.
- Real-World Applications: Logarithms model phenomena involving growth rates, sound intensity (decibels), earthquake magnitude (Richter scale), computer algorithms, and more.
- Foundation for Advanced Topics: Understanding how to solve log equations prepares you for solving complex logarithmic expressions and inequalities in higher-level mathematics.
Quick Tips When Solving log₂(8x) = 5
- Always convert logarithmic equations to exponential form.
- Simplify coefficients—here, 2⁵ = 32.
- Isolate the variable by dividing both sides appropriately.
- Always check solutions by substituting back into the original equation, especially when variables are inside the logarithm.