Sum of Roots

Algebra Level 4

$\large (x+1)=2 \log_2(2^x +3) - 2\log_4(1980-2^{-x})$

If the sum of the roots of the above equation is of the form $\log_4 (k^2)$ then, find the value of $k$ .

The answer is 11.

1 solution

Ashish Gupta
Apr 12, 2016

$x+1=2 \log_2 (2^x + 3) - \log_2 (1980-2^{-x})$

$x+1= \log_2 \left (\frac{(2^x + 3)^2}{1980-2^{-x}} \right )$

$2^{x+1} = \frac{(2^x + 3)^2}{1980-2^{-x}}$

$2= \frac{(2^x + 3)^2}{1980 \cdot 2^x -1}$

Let $2^x = y$ :

$(y+3)^2 = 3960y -2$

$y^2 - 3954y + 11 = 0$

We need $x_1 + x_2$ :

$y_1 \cdot y_2 = 2^{x_1 + x_2} = 11$

$x_1 + x_2 = log_2 (11) = log_4 ({11}^2)$

Hence, $k=11$

Very false!!! $log_{2}(11) = 2log_{4}(11) = log_{4}(121) = log_{4}(\sqrt{14641})$

Therfore the correct answer should be $\boxed{14641}$ .

Andreas Wendler - 5 years, 2 months ago

Those who answered 14641 have been marked correct.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the menu. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 5 years, 2 months ago

Thank you. Since I now know how to find this special menu I will of course apply it in future if necessary.

Andreas Wendler - 5 years, 2 months ago

Sorry, my bad. Corrected it.

Ashish Gupta - 5 years, 2 months ago

Sum of Roots

The answer is 11.

1 solution

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