How do you solve #5^(x+2) = 4^(1-x)#?

2 Answers
May 22, 2015

One easy way is using logarithms because there is one law of logarithms which states that #loga^n=n*loga#

So, let's put logarithms on both sides (to keep the equality):

#log5^(x+2)=log4^(1-x)#
#(x+2)log5=(1-x)log4#

Approximating #log5~=0.7# and #log4~=0.6#, we get

#(x+2)0.7=(1-x)0.6#
#0.7x+1.4=0.6-0.6x#
#1.3x=-0.8#
#x=-0.8/1.3~=-0.61#

May 22, 2015

Another way to do that would be
#5^(x+2)=4^(1-x)#
#5^x cdot 5^2=4^1 cdot 4^(-1)#
#25 cdot 5^x = 4 cdot 1/4^x#
#5^x cdot 4^x=4/25#
#20^x=0.16#
using the natural log
#log 20^x=log0.16#
#x log 20=log 0.16#
#x=(log 0.16)/(log 20) approx -0.612#
or using the log base 20
#log_(20) 20^x=log_(20) 0.16#
#x=log_(20) 0.16 approx -0.612#