Question

6^2x+4 = 3^3x . 2^x+8. Solve.... Plz

Answer 1

6^2x+4 = (3^3x )( 2^x+8)

Answer 2

\[6^{2x+4} = (3^3x)(2^{x+8})\]

Answer 3

yup! n BTW this related with inequalities !

Answer 4

help me fast plzz!

Answer 5

(2x+4)log 6=(27x) 2^(x+8)
(2x+4)log 6= log 27x + (x+8)log 2

Sorry, but I can't figure it out. Repost it and get some more help.

Answer 6

@dumbcow, can you do this for dyanesh?

Answer 7

Is it 3^2 * x or 3^(2x)?

Answer 8

@radar has the right idea, im assuming thats a typo and is should be 3^(3x) with x in the exponent
if x is not an exponent then it can't be solved algebraically

i will continue where radar left off
\[(2x+4)\log 6 = x \log 27 + (x+8)\log 2\]
\[2x \log 6 -x \log 27 -x \log 2 = 8 \log 2 -4 \log 6\]
\[x (2 \log 6 -\log 27-\log 2) = 8 \log 2 -4 \log6\]
\[x = \frac{8 \log2-4 \log6}{2 \log 6 -\log 27-\log 2} = 4\]

Answer 9

Thanks @dumbcow, you made it look simple.