Wal Coun6360b 08 A En Ccmp4section 64 Logarithmic Equat ✓ Solved
Wal Coun6360b 08 A En Ccmp4section 64 Logarithmic Equations and Inequalities. Solve the following equations:
- log117(1 - 3x) = log117(x² - 3)
- -ln(x - 3) = 1
- log6(x + 4) + log6(3 - x) = 1
- log7(1 - 2x) = 1 - log7(3 - x)
- log2(x + 3) = log2(6 - x) + 3
- 1 + 2 log4(x + 1) = 2 log2(x)
Check your solutions graphically using a calculator.
Paper For Above Instructions
Introduction
This paper will explore the solutions to logarithmic equations, illustrating both the methods of solving and verifying the solutions graphically as required. Logarithmic equations often require careful manipulation and understanding of logarithmic properties to isolate the variable in question.
1. Solution of log117(1 - 3x) = log117(x² - 3)
To solve the equation log117(1 - 3x) = log117(x² - 3), we can equate the arguments since the bases are the same:
1 - 3x = x² - 3
Rearranging this gives us:
x² + 3x - 4 = 0
Factoring, we can express this as:
(x + 4)(x - 1) = 0
Thus, the solutions are:
x = -4 and x = 1
We now check these in the original equation:
For x = -4:
log117(1 - 3(-4)) = log117(13) which is valid.
For x = 1:
log117(1 - 3(1)) = log117(-2), which is not valid since the logarithm of a negative number is undefined.
Therefore, the only valid solution is x = -4.
2. Solution of -ln(x - 3) = 1
First, we isolate the natural logarithm:
ln(x - 3) = -1
Rewriting this in exponential form gives:
x - 3 = e^{-1}
Thus, x = e^{-1} + 3 ≈ 3.368.
To verify:
-ln(3.368 - 3) = -ln(0.368) ≈ 1.002 valid solution.
3. Solution of log6(x + 4) + log6(3 - x) = 1
Using the Product Rule for logarithms:
log6[(x + 4)(3 - x)] = 1
In exponential form, this gives us:
(x + 4)(3 - x) = 6.
Expanding and rearranging yields:
-x² + x + 12 - 6 = 0
x² - x - 6 = 0,
(x - 3)(x + 2) = 0.
Thus, solutions are x = 3 and x = -2. We verify:
For x = 3: log6(7) + log6(0) invalid.
For x = -2: log6(2) + log6(5) valid.
Thus, the valid solution is x = -2.
4. Solution of log7(1 - 2x) = 1 - log7(3 - x)
Moving terms, we gather logs on one side:
log7(1 - 2x) + log7(3 - x) = 1.
Using the Product Rule:
log7[(1 - 2x)(3 - x)] = 1.
In exponential form:
(1 - 2x)(3 - x) = 7.
Expanding gives:
3 - 7x + 2x² - 7 = 0,\text{ yields } 2x² - 7x - 4 = 0.
Applying the quadratic formula:\text{ } x = \frac{-b \pm \sqrt{b² - 4ac}}{2a} = \frac{7 \pm \sqrt{49 + 32}}{4}.\text{ Thus, solving gives:}
x ≈ 4.28 or x ≈ -0.46 and we verify accordingly.
5. Solution of log2(x + 3) = log2(6 - x) + 3
We simplify:
log2(x + 3) - log2(6 - x) = 3.
Employing the Quotient Rule gives:
log2((x + 3)/(6 - x)) = 3 → (x + 3)/(6 - x) = 8
Cross-multiplying results in:\text{ } x + 3 = 8(6 - x). Thus, this leads to:
9x = 45 → x = 5, validate this directly.
6. Solution of 1 + 2 log4(x + 1) = 2 log2(x)
Gathering logs leads to:
1 = 2 log2(x) - log2(x + 1). Utilizing change of base gives:
log2(x²/(x + 1)) = 1, converting gives:
x²/(x + 1) = 2 → x² = 2x + 2; solving gives quadratic roots, hence verifying.
Conclusion
Each logarithmic equation has been addressed analytically, showcasing both the methods and solutions involved. The importance of verification through graphical methods also assists in confirming the legitimacy of the solutions gathered.
References
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