Grade 11 · Logarithms and domains · 48 of 600
Logarithms and domains · Domain boundary a=0.5; Logarithm base b=4
Use x=3.58; the logarithm argument is exactly 3.08. State the domain, evaluate the logarithm, and recover the argument by exponentiation. Givens: Domain boundary a=0.5; Logarithm base b=4.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Let f(x)=log_4(x−(0.5)). The base is 4>1. Use x=3.58; the logarithm argument is exactly 3.08. State the domain, evaluate the logarithm, and recover the argument by exponentiation.
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A starting point
Check x−a>0 first, then use ln(x−a)/ln(b). Exponentiating returns the argument before adding the shift.
Work through the reasoning
Step 1
Identify the model and target
Let f(x)=log_4(x−(0.5)). The base is 4>1. Use x=3.58; the logarithm argument is exactly 3.08. The governing relation is y=log_b(x−a); bʸ=x−a; x>a. Check x−a>0 first, then use ln(x−a)/ln(b). Exponentiating returns the argument before adding the shift.
Step 2
Substitute and calculate
The argument is 3.58−(0.5)=3.08. Change of base gives ln(3.08)/ln(4)=0.811465; raising 4 to this output recovers 3.08.
Step 3
Check the mathematical meaning
The domain is x>0.5, and 3.58>0.5. The inverse check is 4^(0.811465)≈3.08; adding 0.5 recovers x=3.58. Allowed input x: 3.58; Logarithm value: 0.811; Base raised to output: 3.08. Decimal values are rounded, so use unrounded intermediate values.
The answer
The argument is 3.58−(0.5)=3.08. Change of base gives ln(3.08)/ln(4)=0.811465; raising 4 to this output recovers 3.08. The domain is x>0.5, and 3.58>0.5. The inverse check is 4^(0.811465)≈3.08; adding 0.5 recovers x=3.58. Animation check: Allowed input x: 3.58; Logarithm value: 0.811; Base raised to output: 3.08. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
A real logarithm requires a positive argument. The shift a moves that boundary, while a base greater than one controls how quickly output grows. Exponentiating the output recovers the positive argument, not the original x until the shift is added back. Let f(x)=log_4(x−(0.5)). The base is 4>1. Use x=3.58; the logarithm argument is exactly 3.08. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
y=log_b(x−a); bʸ=x−a; x>a
03 · Reflect and transfer
Explain what changes and why.
With the argument fixed, why does increasing a base greater than one decrease this positive logarithm?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.