Example Question - exponential equation
Here are examples of questions we've helped users solve.
Exponential Equation Involving Base 5 and 25
<p>Given:</p> <p>\(25^{x-1}=5^{2x-1}-100\)</p> <p>Since \(25\) is a power of \(5\), i.e., \(25 = 5^2\), rewrite the equation:</p> <p>\((5^2)^{x-1}=5^{2x-1}-100\)</p> <p>Apply the exponent product rule \( (a^m)^n = a^{mn} \):</p> <p>\(5^{2(x-1)}=5^{2x-1}-100\)</p> <p>\(5^{2x-2}=5^{2x-1}-100\)</p> <p>Now, set the exponents of base \(5\) equal to each other and solve the resulting equation:</p> <p>\(2x-2=2x-1\)</p> <p>The above equation has no solution for \(x\) since \(2x-2\) cannot equal \(2x-1\). However, since it's a proof that without subtracting the \(100\), the equation would never hold true for any value of \(x\), we may have made an error. Instead, let's look for a different approach to this problem:</p> <p>Let \(y=5^{x-1}\), this gives us the following equation: </p> <p>\(y = 5y - 100\)</p> <p>Solve for \(y\):</p> <p>\(100 = 5y - y\)</p> <p>\(100 = 4y\)</p> <p>\(y = 25\)</p> <p>Substituting back for \(y\):</p> <p>\(5^{x-1} = 25\)</p> <p>Since \(25=5^2\), we have:</p> <p>\(5^{x-1} = 5^2\)</p> <p>Thus, \(x-1 = 2\)</p> <p>Finally, solve for \(x\):</p> <p>\(x = 2 + 1\)</p> <p>\(x = 3\)</p>
Solving an Exponential Equation with Fractions
题目给出了一个方程：\[3^a - 5^b = m\]，以及 \[\frac{1}{a} + \frac{1}{b} = 2\]，让我们求 \(m\) 的值。 首先来处理第二个方程 \[\frac{1}{a} + \frac{1}{b} = 2\]。我们寻找能够使这个方程成立的 \(a\) 和 \(b\) 的值。因为这个方程涉及的是两个变数的倒数的和，所以可以预想 \(a\) 和 \(b\) 可能是比较小的整数。让我们尝试一些可能的整数对： - \(a = 1, b = 1\)，那么 \(\frac{1}{a} + \frac{1}{b} = 1 + 1 = 2\)，满足条件，但这会导致 \(b\) 的值为 1，从而 \(5^b = 5\)，这似乎会让第一个方程变得没有意义，因为我们通常不考虑负数作为幂的底数，所以这组解不符合常规的数学规则。 - \(a = 2, b = 2\)，同样 \(\frac{1}{a} + \frac{1}{b} = \frac{1}{2} + \frac{1}{2} = 1\)，并不满足条件。 - \(a = 1, b = 2\) 或者 \(a = 2, b = 1\)，都可以使得 \(\frac{1}{a} + \frac{1}{b} = 1 + \frac{1}{2} = \frac{3}{2}\)，也不满足条件。 继续尝试其他整数对，我们发现 \(a = 1, b = 2\) 或者 \(a = 2, b = 1\) 是错误的选择。但如果 \(a = 1, b = -2\) 或者 \(a = -2, b = 1\)，就会满足方程 \[\frac{1}{a} + \frac{1}{b} = 2\]（由于是求和，所以负值也可以考虑进来）。 现在处理第一个方程 \[3^a - 5^b = m\]。 若使用 \(a = 1, b = -2\)，那么 \(3^a = 3\) 和 \(5^b = 5^{-2} = \frac{1}{25}\)，从而： \[m = 3 - \frac{1}{25} = \frac{75}{25} - \frac{1}{25} = \frac{74}{25}\] 因此 \(m\) 的值是 \[\frac{74}{25}\]。
Solving Exponential Equation with Logarithms
The equation provided in the image is: 5^(3x) + 2 = 8 To solve for x, you should follow these steps: 1. Subtract 2 from both sides to isolate the exponential term: 5^(3x) + 2 - 2 = 8 - 2 5^(3x) = 6 2. Take the logarithm of both sides of the equation. It's often easier to use the logarithm with the same base as the exponent, but you can use the natural logarithm (ln) or common logarithm (log base 10): ln(5^(3x)) = ln(6) 3. Use the power rule of logarithms, which states that ln(a^b) = b*ln(a): 3x * ln(5) = ln(6) 4. Divide both sides by 3*ln(5) to solve for x: x = ln(6) / (3 * ln(5)) Now, we can compute the value of x numerically. Note that since you're asked to round to the nearest thousandth, you should use a calculator to ensure accuracy. x ≈ ln(6) / (3 * ln(5)) Using a calculator: x ≈ ln(6) / (3 * ln(5)) x ≈ 1.791759469 / (3 * 1.609437912) x ≈ 1.791759469 / 4.828313737 x ≈ 0.371072 Rounding to the nearest thousandth: x ≈ 0.371 So, x is approximately 0.371 when rounded to the nearest thousandth.
Solving a Basic Exponential Equation
Dựa vào bức ảnh bạn cung cấp, chúng ta có một bài toán phương trình cơ bản với phần mô tả bằng tiếng Việt như sau: "Tập nghiệm của biểu thức phương trình \(2^{x-1} = 7 \cdot 2^x + 15\) là:". Để giải phương trình này, chúng ta cần biến đổi nó về dạng cơ bản hơn để tìm giá trị của x. Bắt đầu bằng cách viết lại phương trình đã cho: \(2^{x-1} = 7 \cdot 2^x + 15\) Lưu ý rằng: \(2^{x-1} = \frac{2^x}{2}\) Đưa phương trình về cùng cơ số 2 để thuận tiện trong việc giải: \(\frac{2^x}{2} = 7 \cdot 2^x + 15\) Nhân cả hai vế phương trình cho 2 để loại bỏ mẫu số: \(2^x = 14 \cdot 2^x + 30\) Đưa vế với số mũ x về cùng một bên: \(2^x - 14 \cdot 2^x = 30\) Ta được: \(-13 \cdot 2^x = 30\) Chia cả hai vế cho -13: \(2^x = \frac{30}{-13}\) Phương trình này không có nghiệm thực vì 2 mũ x luôn luôn dương, trong khi \( \frac{30}{-13} \) là số âm. Vậy phương trình không có nghiệm thực, và tập nghiệm là rỗng. Nhìn vào các phương án lựa chọn trong hình ảnh, tập nghiệm của biểu thức phương trình \(2^{x-1} = 7 \cdot 2^x + 15\) là: A) Rỗng. Nên, đáp án đúng là A.
Solving an Exponential Equation with Base 2
To solve the equation \(4^{3x} = 32^{x-5}\), we can express both sides of the equation with a base of 2, as both 4 and 32 are powers of 2: \(4 = 2^2\) and \(32 = 2^5\). So, we can rewrite the equation as \((2^2)^{3x} = (2^5)^{x-5}\). Using the exponentiation rule \( (a^b)^c = a^{bc} \), we get \(2^{6x} = 2^{5x - 25}\). Since the bases are the same, we can set the exponents equal to each other: \(6x = 5x - 25\). Now, we solve for \(x\): \(6x - 5x = -25\) \(x = -25\). Thus, the solution to the equation is \(x = -25\).
Solving an Exponential Equation
The equation in the image is: \(2^{3x+5} = 2^{1-x}\) To solve for \(x\), since the bases are the same, you can set the exponents equal to each other: \(3x + 5 = 1 - x\) Now, solve for \(x\): Add \(x\) to both sides: \(3x + x + 5 = 1 - x + x\) \(4x + 5 = 1\) Subtract 5 from both sides: \(4x + 5 - 5 = 1 - 5\) \(4x = -4\) Divide both sides by 4: \(x = -4 / 4\) \(x = -1\)
Solving Exponential Equation with Substitution
The image shows an equation: \[e^{2x} - 4 \cdot 2^x + 4 = 0\] To solve this equation, let's apply a substitution. Notice that the terms \(e^{2x}\) and \(2^x\) suggest that we can set \(u = 2^x\). Then \(e^{2x} = (e^x)^2 = (2^x)^2 = u^2\). Our equation now becomes: \[u^2 - 4u + 4 = 0\] Now, factor this quadratic equation: \[(u - 2)^2 = 0\] Solving for \(u\), we get \(u = 2\). Since we set \(u = 2^x\), we now have: \[2^x = 2\] Taking the logarithm base 2 of both sides gives: \[x \cdot \log_2(2) = \log_2(2)\] \[\Rightarrow x = 1\] So, the solution to the equation is \(x = 1\).
Solving Exponential Equation $2^x + 3^x = 4^x$
The image you've provided shows an equation: \( 2^x + 3^x = 4^x \). To solve this equation for \( x \), we can try to look for integer solutions by inspection or systematically, assuming \( x \) could be a real number. First, let's examine the behavior of each term as \( x \) increases: - \( 2^x \) and \( 3^x \) both grow at an exponential rate, but \( 3^x \) grows more quickly than \( 2^x \) because it has a larger base. - \( 4^x \) also grows at an exponential rate, even more quickly than both \( 2^x \) and \( 3^x \), since \( 4 = 2^2 \). Given the nature of the equation, we can see that as \( x \) becomes large, \( 4^x \) will dominate the left side of the equation making \( 2^x + 3^x \) less than \( 4^x \). Now, let's consider small values of \( x \): - When \( x = 0 \), \( 2^x + 3^x = 4^x \) because each term equals 1 (any nonzero number to the zeroth power is 1). - When \( x = 1 \), \( 2^x + 3^x = 4^x \) because \( 2 + 3 = 5 \) which is not equal to \( 4 \). - When \( x = 2 \), \( 2^x + 3^x = 4^x \) because \( 4 + 9 = 13 \) which is not equal to \( 16 \). The case when \( x = 0 \) is the only integer solution that satisfies the equation since for \( x > 0 \), the left side of the equation will always be smaller than the right side. Let's try to confirm this by solving the equation algebraically. If we did not find any solutions through inspection or want to consider non-integer real number solutions, we'd have to solve the equation algebraically. However, because this equation is transcendental (involves variable exponents across terms that cannot be easily equated), finding such solutions usually requires numerical methods or advanced mathematical techniques beyond primary algebra. In typical pre-calculus or calculus classes, if this problem was given, the teacher would expect the students to find integer solutions through inspection, as described. For non-integer solutions, students would generally be instructed to use a graphing calculator or numerical methods to approximate solutions, but algebraic methods for finding these approximate solutions without a calculator are largely impractical for this form of the equation. As concluded from inspection, the only solution that can easily be found is \( x = 0 \).
Solving Exponential Equation with Natural Logarithm
To solve the equation \( e^{4 - 7x} + 11 = 20 \), you first isolate the exponential part: 1. Subtract 11 from both sides to move the constant term to the right-hand side of the equation: \[ e^{4 - 7x} = 20 - 11 \] \[ e^{4 - 7x} = 9 \] 2. Now take the natural logarithm (ln) of both sides to solve for \( 4 - 7x \): \[ \ln(e^{4 - 7x}) = \ln(9) \] Since the natural logarithm and the exponential function are inverse functions, \( \ln(e^{y}) = y \), you get: \[ 4 - 7x = \ln(9) \] 3. Finally, solve for x: \[ 7x = 4 - \ln(9) \] \[ x = \frac{4 - \ln(9)}{7} \] This is the solution to the equation.
Solving Exponential Equation for x
The equation provided in the image is: \[ e^{4-7x} + 11 = 20 \] To solve for \( x \), we'll follow these steps: 1. Subtract 11 from both sides of the equation. \[ e^{4-7x} = 9 \] 2. Take the natural logarithm (ln) of both sides to get rid of the base \( e \). \[ \ln(e^{4-7x}) = \ln(9) \] 3. Use the property of logarithms that \( \ln(e^y) = y \) to simplify the left side. \[ 4 - 7x = \ln(9) \] 4. Subtract 4 from both sides to isolate the term with \( x \). \[ -7x = \ln(9) - 4 \] 5. Finally, divide both sides by -7 to solve for \( x \). \[ x = \frac{\ln(9) - 4}{-7} \] 6. Now you can either leave the answer in terms of natural logarithm or use a calculator to find a numerical value for \( x \). Using a calculator: \[ x \approx \frac{2.1972 - 4}{-7} \] \[ x \approx \frac{-1.8028}{-7} \] \[ x \approx 0.25754 \] So the solution to the equation is \( x \approx 0.25754 \).
Solving Exponential Equation for x
The equation provided is: \[ e^{4 - 7x} + 11 = 20 \] To solve for \( x \), follow these steps: 1. Subtract 11 from both sides of the equation: \[ e^{4 - 7x} = 9 \] 2. Take the natural logarithm (ln) of both sides: \[ \ln(e^{4 - 7x}) = \ln(9) \] 3. Because the natural logarithm and the exponential function are inverse operations, \( \ln(e^{y}) = y \). Hence: \[ 4 - 7x = \ln(9) \] 4. To isolate \( x \), subtract 4 from both sides: \[ -7x = \ln(9) - 4 \] 5. Finally, divide both sides by -7 to solve for \( x \): \[ x = \frac{\ln(9) - 4}{-7} \] So, \[ x = \frac{4 - \ln(9)}{7} \] Now you can use a calculator to find the approximate value of \( x \).
Solving an Exponential Equation
This is an equation involving an exponential function. To solve the equation \(e^{4 - 7x} + 11 = 20\), we need to isolate the term with the exponent and then solve for \(x\). Here are the steps to solve the equation: 1. Subtract 11 from both sides to isolate the exponential term: \(e^{4 - 7x} = 9\) 2. Take the natural logarithm (ln) of both sides to remove the base \(e\) and get the exponent by itself: \( \ln(e^{4 - 7x}) = \ln(9) \) 3. The natural logarithm and the exponential function are inverse functions, so: \(4 - 7x = \ln(9)\) 4. Solve for \(x\): \(7x = 4 - \ln(9)\) \(x = \dfrac{4 - \ln(9)}{7}\) 5. To find the numeric value, we can use a calculator: The natural logarithm of 9 (to two decimal places) is approximately 2.20. \(x \approx \dfrac{4 - 2.20}{7}\) \(x \approx \dfrac{1.80}{7}\) \(x \approx 0.2571\) So the approximate solution to the original equation is \(x \approx 0.2571\).
Solving Exponential Equation
The image shows the following exponential equation: e^(4 - 7x) + 11 = 20 To solve the equation for x, we'll follow these steps: 1. First, isolate the exponential term: e^(4 - 7x) = 20 - 11 e^(4 - 7x) = 9 2. Next, we would take the natural logarithm (ln) of both sides to get the exponent by itself: ln(e^(4 - 7x)) = ln(9) 3. Since ln(e^y) = y for any y (because ln and e are inverse functions), we can simplify the left-hand side: 4 - 7x = ln(9) 4. Lastly, we'll solve for x: 7x = 4 - ln(9) x = (4 - ln(9)) / 7 To find the numerical value, compute: ln(9) ≈ 2.1972 So, x ≈ (4 - 2.1972) / 7 x ≈ 1.8028 / 7 x ≈ 0.2575 Therefore, the solution to the equation is x ≈ 0.2575.
Solving an Exponential Equation with Two Variables
The image shows an exponential equation: e^(t + 7x) + 11 = 20. To solve for the variables t and x in this equation, follow these steps: 1. Isolate the exponential term on one side of the equation: e^(t + 7x) = 20 - 11 e^(t + 7x) = 9 2. Since e^(t + 7x) = 9, take the natural logarithm of both sides to get rid of the exponential base 'e': ln(e^(t + 7x)) = ln(9) 3. By the property of logarithms, ln(e^y) = y for any y, we have: t + 7x = ln(9) From this point, the equation cannot be solved for a unique solution because we have a single equation with two variables. To find specific values for t and x, we would need a second independent equation. If there is additional information or another equation provided, you can use it to solve for the unknowns. Otherwise, you have an infinite set of solutions depending on the value of one of the variables.
Solving Exponential Equation with Natural Logarithm
The equation in the image is \( e^x + 11 = 20 \). To solve for \( x \), you want to isolate \( e^x \) on one side of the equation. Here are the steps to do that: 1. Subtract 11 from both sides of the equation to isolate the exponential term on one side: \( e^x + 11 - 11 = 20 - 11 \) Simplifying that, you get: \( e^x = 9 \) 2. Taking the natural logarithm (ln) of both sides to solve for \( x \): \( \ln(e^x) = \ln(9) \) Since \( \ln(e^x) \) simplifies to \( x \)(because \( \ln(e^x) \) and \( e^x \) are inverse functions), you have: \( x = \ln(9) \) The approximate value of \( x \) can be found using a calculator: \( x \approx \ln(9) \approx 2.1972 \)
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