Solve exponential and logarithmic equations

For the displayed task, what is the first algebraic decision?

Test the claim that positive endpoint values prevent interior roots. Inspect f(x) = (x − a)² − b² over [a − 2b,a + 2b]. Change a or b while keeping b positive. This table gives exact theoretical values

Why is every displayed setting a counterexample to the claim?

Solving exponential and logarithmic equations

Isolate an exponential expression or combine logarithms before solving for the variable. Apply the same logarithm to both sides, check every original logarithm argument, and round only the final approximation

1. 1Isolate the exponential Make the exponential expression the subject before applying logarithms \(\displaystyle 3\times5^{x-1}=18\Rightarrow5^{x-1}=6\)
2. 2Apply the same logarithm Apply one logarithm base to both sides so the exponent can be brought down \(\displaystyle 5^x=12\Rightarrow\log_5(5^x)=\log_5(12)\Rightarrow x=\log_5(12)\)
3. 3Convert a logarithm to exponential form A logarithm states the exponent that produces its positive argument \(\displaystyle \log_3(x+1)=2\Rightarrow x+1=3^2=9\)
4. 4Check domains and round once Every original logarithm argument must be positive, and an approximate answer is rounded only at the end \(\displaystyle \log_2(x-3)=4\Rightarrow x=19,\quad x-3=16>0\)

Given

\(\displaystyle 4^{2x+1}=32\Rightarrow2x+1=\log_{4}\left(32\right)\)

Evaluate the logarithm in the rearranged equation

Given

\(\displaystyle 2x+1=\frac{5}{2}\Rightarrow2x=\frac{3}{2}\)

Use the exact logarithm value to solve for x.

Taking a logarithm on one side. One of these lines is right. Which one?

Keep an actual technology record

Use technology with graphical, numerical and symbolic functions. Keep the entered expression, assumptions, selected domain and viewing window, actual output, independent hand check and explanation

1. 1Use the functions the task needs Use a CAS for the symbolic solving step, not just a numerical solver or a graph. If your tool has no symbolic facility, leave that execution step pending until a suitable tool is available. Hand algebra verifies an output; it does not stand in for a recorded technology run
2. 2Separate three kinds of bounds The mathematical domain determines allowed inputs. A deliberate domain restriction changes the problem. A viewing window only controls the portion displayed and may conceal valid features. Record the displayed horizontal and vertical limits separately

Worked model: refute a root claim

Claim: a continuous function with positive values at both ends of an interval has no roots inside. A counterexample must meet the stated conditions and contradict the conclusion. Use f(x) = (x − 1/2)² − 1 on [−3/2,5/2]

1. 1Plan the technology record Enter (x-0.5)^2-1 with explicit brackets. Use x from −1.5 to 2.5 and y from −1.5 to 3.5. Request a table at −1.5, −0.5, 0.5, 1.5 and 2.5. These settings reveal the endpoints, vertex and crossings
2. 2Expected output to check The corresponding exact values are 3, 0, −1, 0, 3. This is an authored prediction, not a claim that a particular device was run
3. 3Model explanation The polynomial is continuous, and both endpoint values are 3 > 0. Yet its roots are −1/2 and 3/2, both strictly inside the interval. Therefore the claim is false. The graph helps find the example; exact substitution verifies it

Your task: produce and verify a counterexample

Test the claim that a continuous function with positive values at both ends of an interval has no roots inside. Choose your own a from −1, 0, 1, 2 and b from 0.5, 1, 1.5. Use your calculator, spreadsheet or graphing tool to produce a graph and five-row table for f(x) = (x − a)² − b² on [a − 2b,a + 2b]. Choose a vertical window that shows −b² and 3b². Save three settings, changing only a or only b at a time. Compare their graphs and tables and explain what changes and what stays true. Keep your actual input and output; do not substitute the authored table for your technology record

1. 1Written explanation State the claim, your parameter choices, continuity, both endpoint signs and the two interior roots. Explain why one valid counterexample refutes a universal claim. A graph picture alone is not that explanation
2. 2Independent checks Substitute the endpoints and roots by hand. Check that b is positive, the roots are strictly inside and every table value matches your entered rule. If the output differs, investigate brackets or settings before concluding

Given

\(\displaystyle f(x)=(x+1)^2-9/4,\quad x=-4\)

Separate check: for a = −1 and b = 3/2, calculate f at the left endpoint a − 2b

Given

\(\displaystyle a=-1,\quad b=3/2\)

For this same fixed example, calculate the smaller root a − b.

Worked model: preserve meaning across notation

Formal mathematics writes r(x) = (x² − 16)/(x − 4), with x ≠ 4. A common linear input is (x^2-16)/(x-4). The ^ denotes a power and the brackets group the entire numerator and denominator. Your device may require different syntax; record its actual input and output

1. 1Simplified output and original domain Factoring gives (x − 4)(x + 4)/(x − 4) = x + 4 only for x ≠ 4. A simplifier may display x + 4 without the exclusion. The original expression remains undefined at 4; the displayed polynomial alone would give 8 there
2. 2An output convention worth checking For real x, sqrt(x^2) is |x|, often written abs(x), not always x. At x = −3 the value is 3. With an assumption x ≥ 0 a tool can legitimately simplify it to x; record assumptions as well as output
3. 3Model comparison At x = 2 both admissible forms of r give 6, but agreement at one input does not prove identity. Factorisation proves agreement on the original domain. Domain restrictions and real-variable assumptions are part of the statement

Your task: translate and compare technology output

Choose a = 2, 3 or 5 and enter (x² − a²)/(x − a) into technology using its supported syntax. Use symbolic simplification and record its actual output, then compare a table of the original and candidate x + a at admissible inputs and at x = a. Also inspect sqrt(x²) at a negative and positive input

1. 1Make a three-column record Write the formal expression and domain, your exact technology input/settings, and the actual output. Explain each power, bracket, absolute-value notation and any undefined/error result. Record real-variable assumptions or restrictions you supplied
2. 2Model for the reasoning The two rational forms agree when x ≠ a because the common factor can be cancelled there. The polynomial extends to a value at x = a, while the original does not. For sqrt(x²), a negative test distinguishes |x| from x; factorisation or a sign argument completes the explanation

Given

\(\displaystyle r(x)=\frac{x^2-25}{x-5},\quad x=-2\)

Separate check: evaluate the original rational expression at x = −2.

For r(x) = (x² − 25)/(x − 5), a tool displays x + 5. What must you preserve?

Given

\(\displaystyle \sqrt{x^2},\quad x=-7/2\)

Evaluate the principal square root at this negative input

Worked plan: exact roots, decimal values and a graph

Take f(x)=log₂((x−1)²) and solve f(x)=1. The maximal real domain excludes x=1. A common linear input is ln((x-1)^2)/ln(2); use your tool’s supported syntax. Symbolic solving should give 1−√2 and 1+√2. Numerical evaluation gives approximately −0.4142135624 and 2.4142135624. A graph of f and y=1, with x from −1 to 3 and y from −3 to 3, reveals both intersections and the gap at 1. The table at −1,0,1,2,3 gives 2,0,undefined,0,2. These are authored expected results, not a claim that a device was run

1. 1Why a narrow graph can mislead A horizontal window from 0.5 to 1.5 hides both roots or intersections. That display does not prove they do not exist. Restore the larger window and use the exact symbolic answers and independent substitution to explain the difference
2. 2Certify rounding separately The exact squares show 1.4142135 < √2 < 1.4142136. Translate both endpoints for the larger root and reverse them when subtracting for the smaller root. Both endpoints of each resulting interval round to the same three-decimal value, certifying −0.414 and 2.414. A displayed midpoint or truncated screen alone is not this proof

Rule

\(\displaystyle f(x)=\log_2((x-1)^2)\)

For the worked logarithmic rule, find f(3)

Produce graphical, symbolic and numerical results

Choose a=2,3 or 4 and use f(x)=log₂((x−a)²). State the maximal real domain and solve f(x)=1 using your tool’s symbolic solver. Record its exact output and any restrictions or assumptions. Evaluate every exact solution numerically, retain the unrounded values and report three decimal places. Independently convert the logarithmic equation to an algebraic equation, retain the original domain, and verify every solution

1. 1Choose and justify a useful view Produce a graph of f and y=1 and a table including a−2,a−1,a,a+1,a+2. Choose horizontal and vertical window limits that reveal both intersections, both branches and the excluded input. Save your actual settings/output and explain why they are suitable. Then use the narrow horizontal window [a−0.5,a+0.5] and record which intersections disappear. Finally apply the domain restriction [a−2,a+2] excluding a in the tool and save that graph. State its true range and explain why a finite vertical viewing window cannot display the entire range
2. 2Verify, explain and retain Use 1.4142135 < √2 < 1.4142136 to certify both rounded answers by endpoint rounding, not just by copying the display. Check the entered brackets, variable value, exact solution form and table values. Compare the usefulness and limitations of the three functions you used. Keep the complete run record and your independent reasoning for review

Review every part of the record

For f(x)=log₂((x−a)²), where a is 2, 3 or 4, solving f(x)=1 gives a−√2 and a+√2; neither equals the excluded input a. Their three-decimal values are (0.586,3.414), (1.586,4.414) or (2.586,5.414) for a=2,3,4 respectively. The five-row table is 2,0,undefined,0,2. The narrow window hides both intersections because √2>0.5. The maximal domain is R except a and the maximal range is R. On [a−2,a+2] excluding a, the range is (−∞,2]; a finite display clips its unbounded lower part. Check that both branches and both solutions appear in the saved output

1. 1Compare the actual record with the prediction The displayed numbers and model graph are checks, not evidence of your run. Compare your actual commands, output, assumptions, selected domain and window, and independent proof. Investigate missing branches or roots before accepting the result

A logarithm rewrite can lose half the domain

The claim log₂((x−a)²)=2log₂(x−a) for every x≠a is false over the reals. For x=a−1 the left side is log₂1=0 but the right side has a negative argument and is undefined. The correct real identity is log₂((x−a)²)=2log₂|x−a| for x≠a. A tool that assumes x>a may legitimately display 2log₂(x−a); that conditional output is not a whole-domain identity

Produce a counterexample and explain the notation

Choose a = 2, 3 or 4 and ask symbolic technology to simplify log₂((x−a)²) and compare it with 2log₂(x−a) and 2log₂|x−a|. Record the actual input syntax, real-variable assumptions and exact output. Produce a table at a−1 and a+1 and use it to refute the claimed whole-domain equality with 2log₂(x−a). Explain how parentheses, squared arguments, absolute-value notation and domain conditions differ between formal mathematics and the device display. If a default complex mode returns a complex value, record that setting and repeat in a real-domain context; do not call the complex expression a real function value

1. 1Explain the missing solution Solve the original equation and the incorrectly restricted rewrite equal to 1. The original has two real solutions; the rewrite with log₂(x−a) allows only the larger one. Use this domain difference and exact substitution to explain the actual output rather than treating one solver line as a complete proof

Review the counterexample and domain conditions

At a−1 the original and absolute-value forms both equal 0; 2log₂(x−a) is not real-defined. At a+1 all three equal 0. One valid counterexample disproves the universal real-domain claim, while agreement at a positive input does not prove it. The original equation log₂((x−a)²)=1 gives a±√2. The restricted rewrite 2log₂(x−a)=1 gives only a+√2, because x>a was imposed. Keep the original domain and any assumptions beside the tool’s equivalent form