What is x if #log_3 (2x-1) = 2 + log_3 (x-4)#?
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To find the value of (x) in the equation (\log_3(2x - 1) = 2 + \log_3(x - 4)), you can use properties of logarithms.
First, apply the product rule of logarithms, which states that (\log_b(M) + \log_b(N) = \log_b(M \cdot N)).
Then, apply the power rule of logarithms, which states that (n \cdot \log_b(M) = \log_b(M^n)).
After combining the logarithms, solve for (x).
[ \begin{align*} \log_3(2x - 1) &= 2 + \log_3(x - 4) \ \log_3(2x - 1) &= \log_3(3^2) + \log_3(x - 4) \ \log_3(2x - 1) &= \log_3(9) + \log_3(x - 4) \ \log_3(2x - 1) &= \log_3(9(x - 4)) \ 2x - 1 &= 9(x - 4) \ 2x - 1 &= 9x - 36 \ -1 + 36 &= 9x - 2x \ 35 &= 7x \ x &= \frac{35}{7} \ x &= 5 \end{align*} ]
So, (x = 5).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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