How do you solve #5(4^36)=4^x#?

Question

1278 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T14:09:42

#x=37.161#

#5(4^36)=4^x# means

#x=log_4(5(4^36)# or

#x=log_4(5)+log_4(4^36)# or

#x=log_4(5)+36log_4(4)# or

#x=log_4 5+36# or

#x=log5/log4+36=1.161+36=37.161#

Answer 2 · 2016-05-09T07:57:33

Use laws of indices first, then logs.
#37.61 = x#

There are powers of 4 on both sides of the equation.

#5(4^36) = 4^x " divide by" 4^36#

#5 = (4^x)/4^36" subtract indices"#

#5 = 4^(x - 36)#

#log 5 = (x - 36)log4#

#log5/log4 = x - 36#

#1.6096 = x - 36#

#37.61 = x#