How do you solve #4(.3)^x = 1.2^(x+2)#?

2 Answers
Jun 21, 2016

# (ln 5 - ln 3)/ (ln 2)#

Explanation:

take logs

#ln ( 4 (0.3)^x) = ln(1.2^{x+2})#
#ln ( 4 ) + ln (0.3^x) = ln(1.2^{x+2})#
#ln ( 4 ) + x ln (0.3) = (x+2) ln(1.2)#
# x (ln (0.3) - ln(1.2) )= 2 ln(1.2) - ln ( 4 )#
# x ln (0.3 / 1.2 )= ln(1.2^2 / 4)#
# x ln (1/4 )= ln(9/25)#
# x = ln(9/25) / ln (1/4 ) = (ln 9 - ln 25)/(ln 1 - ln 4)#
# = (2 ln 3 - 2 ln 5)/(0 - 2 ln 2)#
# = (ln 5 - ln 3)/ (ln 2)#

Jun 21, 2016

#x = 0.736966#

Explanation:

#4(.3)^x = 1.2^(x+2) = (4 xx .3)^{x+2} = 4^{x+2}xx(.3)^{x+2}#

then

#4(.3)^x = 4^{x+2}xx(.3)^{x+2} = 4^2xx(0.3)^2xx4^x xx(.3)^x#

elliminating #(.3)^x# in both sides

#4 = 4^2xx(0.3)^2xx4^x->4^x=1/(4 xx (0.3)^2)#

and finally

#x = -log_e(4 xx (0.3)^2)/log_e 4 = 0.736966#