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- Dec 15th 2009, 10:11 AM #1

- Dec 15th 2009, 11:37 AM #2
1. From the LHS of the equation you know $\displaystyle x = 9^{\log_9(x)}$

2. Plug in this term into the RHS:

$\displaystyle \frac12 \log_3\left({9^{\log_9(x)}}\right) = \frac12 \log_3(9) \cdot \log_9(x) = \frac12 \cdot 2 \cdot \log_9(x) = \log_9(x)$

3. Consequently the given equation

$\displaystyle \log_3(2x+1) = \log_9(5x^2-x+5)$

becomes

$\displaystyle \log_3(2x+1) = \frac12 \log_3(5x^2-x+5)$

$\displaystyle \log_3(2x+1) = \log_3(\sqrt{5x^2-x+5})$

4. Now use both sides as exponents to the base 3:

$\displaystyle 2x+1 = \sqrt{5x^2-x+5}$

Square both sides and solve the quadratic equation for x.

Check afterwards if both solutions are valid.