Question

to solve the equation \\(\log_{8}(x - 3) + \log_{8}(x + 4) = 1\\), one method is to apply the properties of logarithms to write a new equation that can be used to identify possible solutions. of the following, which is such an equation?\
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a \\(2x + 1 = 8\\)\
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b \\(\frac{x - 3}{x + 4} = 8\\)\
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c \\(x^2 - 12 = 8\\)\
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d \\(x^2 + x - 12 = 8\\)

Explanation:

Step1: Combine logs using product rule

$\log_8[(x-3)(x+4)] = 1$

Step2: Convert log to exponential form

$(x-3)(x+4) = 8^1$

Step3: Expand left side

$x^2 + x - 12 = 8$

Answer:

D. $x^2 + x - 12 = 8$