Question

IF the roots of the equation x^2-4x-log3a=0 (log 3base a ) are real , then the least value of a

Answer 1

\[x^2-4x-\log3^a=0\]

Answer 2

@pooja195

Answer 3

I'm going to replace the 'a' with y

hint: if the two roots are real, then D = b^2 - 4ac is going to be greater than or equal to zero. Ie,

b^2 - 4ac >= 0

in this case, a = 1, b = -4, c = log(3,y)

Answer 4

i am getting problem with log case . the way of solving i know .@jim_thompson5910

Answer 5

\[\Large {\color{red}{1}}x^2+({\color{green}{-4}}x)+{\color{blue}{\log_{3}(y)}}\]
is in the form
\[\Large {\color{red}{a}}x^2+{\color{green}{b}}x+{\color{blue}{c}}\]

where
\[\Large \color{red}{a = 1}\]
\[\Large \color{green}{b = -4}\]
\[\Large \color{blue}{c = \log_{3}(y)}\]

Answer 6

\[\Large D \ge 0\]
\[\Large b^2 - 4ac \ge 0\]
\[\Large (-4)^2 - 4(1)(\log_3(y)) \ge 0\]
\[\Large 16 - 4*\log_3(y) \ge 0\]
solve for y

Answer 7

how to exclude log that's my problem

Answer 8

\[\Large 16 - 4*\log_3(y) \ge 0\]
\[\Large 16 \ge 4*\log_3(y)\]
\[\Large 4*\log_3(y) \le 16\]
\[\Large \log_3(y) \le 4\]
\[\Large y \le 3^4\]
\[\Large y \le 81\]

So as long as y is less than or equal to 81, then \(\Large x^2 - 4x + \log_{3}(y) = 0\) will have two real solutions for x.

Answer 9

oh wait, it's not +log(3,y) it's -log(3,y)

so that will change things a bit

Answer 10

a = 1
b = -4
c = -log(3,y)


\[\Large D \ge 0\]
\[\Large b^2 - 4ac \ge 0\]
\[\Large (-4)^2 - 4(1)(-\log_3(y)) \ge 0\]
\[\Large 16 + 4*\log_3(y) \ge 0\]
solve for y

Answer 11

yeah it changed the whole answer

Answer 12

The steps will be very similar

Answer 13

first try to isolate the log itself, then isolate y after that

Answer 14

it asked the least value so will the answer be 81 or 1/81

Answer 15

were you able to solve \[\Large 16 + 4*\log_3(y) \ge 0\]
for y?

Answer 16

y<=81 thanks

Answer 17

no

Answer 18

why

Answer 19

\[\Large 16 + 4*\log_3(y) \ge 0\]
\[\Large 4*\log_3(y) \ge -16\]
\[\Large \log_3(y) \ge -4\]
\[\Large y \ge 3^{-4}\]
\[\Large y \ge \frac{1}{3^4}\]
\[\Large y \ge \frac{1}{81}\]

Answer 20

So as long as y is greater than or equal to \(\Large \frac{1}{81}\), then \(\Large x^2 - 4x - \log_{3}(y) = 0\) will have two real solutions for x.

Answer 21

thanks

Answer 22

np