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Sagot :
To solve the equation [tex]\( 5^{2x} - 20 \cdot 5^{x-1} - 5 = 0 \)[/tex], we can follow these steps:
1. Rewrite the equation in a more manageable form:
Let's make a substitution to simplify the equation. Let [tex]\( y = 5^x \)[/tex]. Then, [tex]\( 5^{2x} \)[/tex] can be written as [tex]\( (5^x)^2 = y^2 \)[/tex], and [tex]\( 5^{x-1} \)[/tex] can be written as [tex]\( \frac{5^x}{5} = \frac{y}{5} \)[/tex].
Substituting these into the equation gives:
[tex]\[ y^2 - 20 \left(\frac{y}{5}\right) - 5 = 0 \][/tex]
2. Simplify the equation:
Simplify the equation by performing the multiplication inside:
[tex]\[ y^2 - 4y - 5 = 0 \][/tex]
3. Solve the quadratic equation:
This is a standard quadratic equation in the form [tex]\( ay^2 + by + c = 0 \)[/tex]. In our case, [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -5 \)[/tex]. We can solve this using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula gives:
[tex]\[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)} \][/tex]
Simplifying inside the square root:
[tex]\[ y = \frac{4 \pm \sqrt{16 + 20}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 6}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4 + 6}{2} = \frac{10}{2} = 5 \][/tex]
[tex]\[ y = \frac{4 - 6}{2} = \frac{-2}{2} = -1 \][/tex]
4. Back-substitute [tex]\( y = 5^x \)[/tex]:
Recall that we let [tex]\( y = 5^x \)[/tex]. Thus, we have:
[tex]\[ 5^x = 5 \][/tex]
[tex]\[ 5^x = -1 \][/tex]
5. Solve for [tex]\( x \)[/tex]:
For [tex]\( 5^x = 5 \)[/tex]:
[tex]\[ x = 1 \][/tex]
For [tex]\( 5^x = -1 \)[/tex]:
[tex]\( 5^x = -1 \)[/tex] has no real solutions since [tex]\( 5^x \)[/tex] is always positive for any real [tex]\( x \)[/tex]. However, in the complex plane, we can use complex logarithms to solve it.
Recall that for any complex number [tex]\( z \)[/tex], [tex]\( z = e^{i\theta} \)[/tex], hence [tex]\( 5^x = -1 \)[/tex] can be written as:
[tex]\[ x \ln(5) = \ln(-1) \][/tex]
The natural logarithm of [tex]\(-1\)[/tex] is [tex]\( i\pi \)[/tex] (considering the principal value), thus:
[tex]\[ x \ln(5) = i\pi \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{i\pi}{\ln(5)} \][/tex]
Therefore, the solutions to the equation [tex]\( 5^{2x} - 20 \cdot 5^{x-1} - 5 = 0 \)[/tex] are:
[tex]\[ x = 1 \quad \text{and} \quad x = \frac{i\pi}{\ln(5)} \][/tex]
1. Rewrite the equation in a more manageable form:
Let's make a substitution to simplify the equation. Let [tex]\( y = 5^x \)[/tex]. Then, [tex]\( 5^{2x} \)[/tex] can be written as [tex]\( (5^x)^2 = y^2 \)[/tex], and [tex]\( 5^{x-1} \)[/tex] can be written as [tex]\( \frac{5^x}{5} = \frac{y}{5} \)[/tex].
Substituting these into the equation gives:
[tex]\[ y^2 - 20 \left(\frac{y}{5}\right) - 5 = 0 \][/tex]
2. Simplify the equation:
Simplify the equation by performing the multiplication inside:
[tex]\[ y^2 - 4y - 5 = 0 \][/tex]
3. Solve the quadratic equation:
This is a standard quadratic equation in the form [tex]\( ay^2 + by + c = 0 \)[/tex]. In our case, [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -5 \)[/tex]. We can solve this using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula gives:
[tex]\[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)} \][/tex]
Simplifying inside the square root:
[tex]\[ y = \frac{4 \pm \sqrt{16 + 20}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 6}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4 + 6}{2} = \frac{10}{2} = 5 \][/tex]
[tex]\[ y = \frac{4 - 6}{2} = \frac{-2}{2} = -1 \][/tex]
4. Back-substitute [tex]\( y = 5^x \)[/tex]:
Recall that we let [tex]\( y = 5^x \)[/tex]. Thus, we have:
[tex]\[ 5^x = 5 \][/tex]
[tex]\[ 5^x = -1 \][/tex]
5. Solve for [tex]\( x \)[/tex]:
For [tex]\( 5^x = 5 \)[/tex]:
[tex]\[ x = 1 \][/tex]
For [tex]\( 5^x = -1 \)[/tex]:
[tex]\( 5^x = -1 \)[/tex] has no real solutions since [tex]\( 5^x \)[/tex] is always positive for any real [tex]\( x \)[/tex]. However, in the complex plane, we can use complex logarithms to solve it.
Recall that for any complex number [tex]\( z \)[/tex], [tex]\( z = e^{i\theta} \)[/tex], hence [tex]\( 5^x = -1 \)[/tex] can be written as:
[tex]\[ x \ln(5) = \ln(-1) \][/tex]
The natural logarithm of [tex]\(-1\)[/tex] is [tex]\( i\pi \)[/tex] (considering the principal value), thus:
[tex]\[ x \ln(5) = i\pi \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{i\pi}{\ln(5)} \][/tex]
Therefore, the solutions to the equation [tex]\( 5^{2x} - 20 \cdot 5^{x-1} - 5 = 0 \)[/tex] are:
[tex]\[ x = 1 \quad \text{and} \quad x = \frac{i\pi}{\ln(5)} \][/tex]
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