Example Question - base 2
Here are examples of questions we've helped users solve.
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 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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