Example Question - exponential form
Here are examples of questions we've helped users solve.
Expressing a Large Number in Scientific Notation
<p>Write \( 450,000,000 \) in scientific notation:</p> <p>\( 450,000,000 = 4.5 \times 10^8 \)</p> <p>Therefore, the floating-point form is:</p> <p>\( 4.5 \) and the exponent is \( 08 \).</p>
Converting Logarithmic Equation to Exponential Form
To write the logarithmic equation log₈ 64 = 2 in exponential form, remember that the logarithmic equation logₐ b = c translates to a^c = b in exponential form. In this case, a = 8, b = 64, and c = 2. Thus, the exponential form of the equation is: 8^2 = 64
Complex Numbers Operations in Polar Form
The image shows a question involving two complex numbers in polar form along with four sub-questions (a to d) asking for the product, quotient, exponential form, and rectangular form of the complex numbers. The complex numbers in question are: \( z = \sqrt{2}(\cos(45^\circ) + i\sin(45^\circ)) \) \( w = 2(\cos(30^\circ) + i\sin(30^\circ)) \) Let's go through each part of the question: a) Write the product \( zw \) in the polar form. The product of two complex numbers in polar form is found by multiplying their magnitudes (r) and adding their angles (θ). For the given complex numbers: \( r_z = \sqrt{2}, \theta_z = 45^\circ \) \( r_w = 2, \theta_w = 30^\circ \) So the product \( zw \) has a magnitude \( r_{zw} = r_z * r_w \) and an angle \( \theta_{zw} = \theta_z + \theta_w \): \( r_{zw} = \sqrt{2} * 2 = 2\sqrt{2} \) \( \theta_{zw} = 45^\circ + 30^\circ = 75^\circ \) Thus, in polar form, the product is: \( zw = 2\sqrt{2}(\cos(75^\circ) + i\sin(75^\circ)) \) b) Write the quotient \( \frac{z}{w} \) in the polar form. The quotient of two complex numbers in polar form is found by dividing their magnitudes and subtracting their angles: \( r_{\frac{z}{w}} = \frac{r_z}{r_w} \) \( \theta_{\frac{z}{w}} = \theta_z - \theta_w \) Therefore: \( r_{\frac{z}{w}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) \( \theta_{\frac{z}{w}} = 45^\circ - 30^\circ = 15^\circ \) In polar form, the quotient is: \( \frac{z}{w} = \frac{\sqrt{2}}{2}(\cos(15^\circ) + i\sin(15^\circ)) \) c) Write \( z \) in the exponential form. To express the complex number \( z \) in exponential form, use Euler's formula: \( re^{i\theta} = r(\cos(\theta) + i\sin(\theta)) \): \( z = \sqrt{2}e^{i45^\circ} \) Note that normally the angle would be in radians, but here we'll leave it in degrees per the format of the question. d) Write \( w \) in the rectangular form. The rectangular (or Cartesian) form of a complex number is \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. Using the given polar form, we calculate: \( a_w = r_w\cos(\theta_w) = 2\cos(30^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \) \( b_w = r_w\sin(\theta_w) = 2\sin(30^\circ) = 2 \times \frac{1}{2} = 1 \) Thus, the rectangular form of \( w \) is: \( w = \sqrt{3} + i \)
Conversion Between Exponential and Logarithmic Forms
The image shows a written question that asks to convert an equation from exponential form to logarithmic form and vice versa. The equation given is: 7^4 = 2401 Let's first convert this exponential form to logarithmic form. The general relationship between exponential and logarithmic forms is: If b^x = y, then log_b(y) = x Applying this to the equation 7^4 = 2401, it converts to: log_7(2401) = 4 This means the logarithm (base 7) of 2401 is equal to 4. Now let's convert from logarithmic form back to exponential form. Using the same relationship backward, the logarithmic form log_7(2401) = 4 converts to: 7^4 = 2401 So we have successfully demonstrated converting the equation from exponential to logarithmic form and back to exponential form.
Solving a Logarithmic Equation with Base 2
The equation provided in the image is a logarithmic equation with base 2. It reads: \[ \log_2(x - 2) + \log_2(8 - x) - \log_2(x - 5) = 3 \] To solve for \( x \), first we can use the logarithmic properties to combine the terms on the left side. The addition of two logarithms with the same base corresponds to the logarithm of the product of the arguments, and the subtraction of two logarithms with the same base corresponds to the logarithm of the quotient of the arguments: \[ \log_2\left(\frac{(x - 2)(8 - x)}{x - 5}\right) = 3 \] Now that we have a single log equation, we can rewrite it in exponential form, with the base raised to the power on the right-hand side equal to the expression inside the log: \[ 2^3 = \frac{(x - 2)(8 - x)}{x - 5} \] From this, we know that: \[ 8 = \frac{(x - 2)(8 - x)}{x - 5} \] Now we can solve for \( x \) by cross-multiplying: \[ 8(x - 5) = (x - 2)(8 - x) \] \[ 8x - 40 = 8x - x^2 - 16 + 2x \] \[ 8x - 40 = - x^2 + 10x - 16 \] Rearrange terms to solve the quadratic equation for \( x \): \[ x^2 - 10x + 24 = 0 \] This is a standard quadratic equation and can be factored as: \[ (x - 4)(x - 6) = 0 \] So the two solutions for \( x \) are: \[ x = 4 \quad \text{and} \quad x = 6 \] However, we need to check these solutions in the original equation to make sure we are not including extraneous solutions and that they make sense in the context of the logarithms (i.e., the arguments of all the logarithms must be positive). Substitute \( x = 4 \): \[ \log_2(4 - 2) + \log_2(8 - 4) - \log_2(4 - 5) \] Since \(\log_2(4 - 5)\) would result in the logarithm of a negative number, which is undefined, \( x = 4 \) is not a valid solution. Substitute \( x = 6 \): \[ \log_2(6 - 2) + \log_2(8 - 6) - \log_2(6 - 5) \] \[ \log_2(4) + \log_2(2) - \log_2(1) \] All the arguments are positive, so this is a valid solution. Since \(\log_2(4) = 2\), \(\log_2(2) = 1\), and \(\log_2(1) = 0\): \[ 2 + 1 - 0 = 3 \] The equation holds true for \( x = 6 \). Thus, the valid solution for the given equation is \( x = 6 \).
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