Suppose a transformed exponential rule is f(x)=3−2·2^(x−1). The base is 2; the minus sign outside the exponential makes the outputs decrease. Before solving for an inverse, build four exact pairs: f(−1)=2.5, f(0)=2, f(1)=1 and f(2)=−1. Swapping each input and output gives candidate inverse pairs (2.5,−1), (2,0), (1,1) and (−1,2).
What restricts the inverse’s inputs? Since 2^(x−1) is positive for every real x, the original output is always less than 3. It can approach 3 but never equal it. As x increases without bound, the output decreases without bound. Therefore f has domain all real numbers and range (−∞,3). Its strict decrease makes each output correspond to exactly one input.
Solid red: f. Dashed navy: its inverse. The equal scales make coordinate reflection honest. Curve ends at the drawing window are not domain endpoints; the curves continue. The marked pairs below distinguish the curves without relying on color.
Point on f
Reflected point on its inverse
(−1,2.5)
(2.5,−1)
(0,2)
(2,0)
(1,1)
(1,1)
(2,−1)
(−1,2)
Predict
Predict the inverse value at 2 and whether the inverse can accept 3. Explain using the table and the range, before manipulating symbols. Also predict its asymptote: reflection swaps the horizontal line y=3 with the vertical line x=3. The inverse’s outputs should cover every real number, because the original inputs did.
A proposed answer log₂(3−y)/2+1 places division outside the logarithm. Check it at y=2 before trusting it. Correctly reversing the order of operations matters; applying a logarithm is not the same as dividing a logarithm afterward.
Experiment
Starting from y=3−2·2^(x−1), isolate the exponential: (3−y)/2=2^(x−1). Its left side must be positive. Taking log₂ gives x−1=log₂((3−y)/2), hence
f⁻¹(y)=1+log₂((3−y)/2), for y<3.
The notation f⁻¹ means inverse function, not the reciprocal 1/f. Substituting y=2 gives 1+log₂(1/2)=0, matching the reflected point. At y=3 the logarithm argument is zero, so the expression is undefined. Above 3 its argument is negative, also outside the real logarithm domain.
Verify both directions, retaining their different domains. For every real x, substitute f(x): the logarithm receives 2^(x−1), and the result is 1+(x−1)=x. Conversely, for y<3, substitute f⁻¹(y) into f: the power of 2 becomes (3−y)/2, giving 3−(3−y)=y. These checks establish an inverse on the stated sets; one convenient numerical pair alone would not.
Understand
Logarithmic inequalities need domain checks before using monotonicity. For log₂(x−1)<log₂(5−x), both arguments must be positive, giving 1<x<5. Because base 2 exceeds 1, its logarithm is increasing, so x−1<5−x, or x<3. Intersecting with the original domain gives 1<x<3. The endpoint 1 is excluded by the logarithm, and 3 by the strict inequality.
For log_(1/2)(x−1)≥−2, first require x>1. The base lies between zero and one, so its logarithm decreases. Since (1/2)^(−2)=4, the inequality becomes x−1≤4, giving 1<x≤5. At x=5 equality holds; at x=2 the logarithm is zero and the inequality holds. Checking x=9 gives −3, which fails. Keeping the inequality direction unchanged would include the wrong side of the boundary.
Log laws also explain graph transformations. On x>0, log₂(4x)=log₂x+2: compressing the input scale by a factor of 4 produces the same graph as translating log₂x upward by 2. Construct both descriptions using this coordinate table, then sketch them on the same axes and compare their points.
x
log₂x
log₂(4x)
log₄x
1/4
−2
0
−1
1
0
2
0
4
2
4
1
Changing the base gives log₄x=(log₂x)/(log₂4)=0.5 log₂x, a vertical scaling, not a horizontal translation. Both base-two and base-four graphs pass through (1,0); the translated graph passes through (1,2). More generally log_b(kx)=log_bx+log_bk requires k>0 and x>0, with b>0 and b≠1. If b is below 1, log_bk changes sign relative to an increasing-base logarithm for the same k>1. Do not transfer a guessed shift direction without evaluating it.
Apply
For a fresh function g(x)=−1+3·(1/2)^(x+2), find the full inverse formula, its domain and range. Check both compositions. Mark the original point at x=−2 and its reflected point. Then independently solve log_(1/3)(2x+1)<−1, showing the original domain and the reversed inequality direction.
Compare your formulas and domain checks
The exponential is positive and decreasing, so g has domain all real numbers and range (−1,∞). Solving (y+1)/3=(1/2)^(x+2) gives g⁻¹(y)=log_(1/2)((y+1)/3)−2, defined for y>−1 with all real outputs. Logarithm and exponential cancellation verifies both compositions on their stated domains. The original point (−2,2) reflects to (2,−2); asymptotes y=−1 and x=−1 are similarly exchanged.
The inequality requires 2x+1>0, or x>−1/2. Decreasing logarithms reverse comparisons, so 2x+1>3 and therefore x>1. This already lies inside the original domain; equality x=1 is excluded because the inequality is strict.
Continue
Keep an inverse record containing the original domain and range, the isolation steps, the inverse’s swapped domain and range, two composition checks, and a reflected coordinate pair. For an inequality, retain the original logarithm domains as well as the final interval. A formula with a correct-looking logarithm but the wrong domain is not a complete inverse or solution.
Connect these algebraic decisions to logarithmic response fitting: an inverse model prediction is meaningful only within its declared mathematical and contextual domains. Its numerical precision and calibration evidence remain separate questions.
Apply · explain · revise
Practice and reasoning
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For f(x)=7+3·5^(2x−4) on all real x, derive the complete inverse formula, state both functions’ domains and ranges, and verify both compositions.
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Let f(x)=2−4·3^(x+1), restricted to x≥−1. Find its range and inverse, including the inverse domain and range. Explain why the inverse of the unrestricted formula has a larger domain.
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Solve log_(1/4)(2x−3)>−1 over the reals. Show the original argument condition and explain the inequality direction when removing the logarithm.
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Solve log₃(x+2)≤log₃(10−2x). Give the original domain before simplifying and test an interior value.
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For x>0 compare g(x)=log₅(25x), h(x)=log₂₅(x), and f(x)=log₅(x). Express g and h using f and describe their graph transformations. Explain why the dilation identity would fail as a real formula if 25 were replaced by −25 while retaining x>0.
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For f(x)=4+2·(1/3)^x, describe its direction, attainable outputs and horizontal asymptote. Derive its inverse and explain the inverse graph’s direction and vertical asymptote without derivatives.
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Reflect and continue
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