How do you solve #4^x * 5^(4x+3) = 10^(2x+3)#?

2 Answers
Mar 28, 2016

#x~~0.65#

Explanation:

#1#. Start by taking the logarithm of both sides, since the bases are not the same on the left and right sides of the equation.

#4^x*5^(4x+3)=10^(2x+3)#

#log(4^x*5^(4x+3))=log(10^(2x+3))#

#2#. Use the log property, #log_color(purple)b(color(red)m*color(blue)n)=log_color(purple)b(color(red)m)+log_color(purple)b(color(blue)n)# to simplify the left side of the equation.

#log(4^x)+log(5^(4x+3))=log(10^(2x+3))#

#3#. Use the log property, #log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)#, to rewrite both sides of the equation.

#xlog(4)+(4x+3)log(5)=(2x+3)log(10)#

#4#. Expand the brackets.

#xlog(4)+4xlog(5)+3log(5)=2xlog(10)+3log(10)#

#5#. Bring all terms with the variable, #x#, to the left side of the equation and all terms without to the right side.

#xlog(4)+4xlog(5)-2xlog(10)=3log(10)-3log(5)#

#6#. Factor out #x# from the terms on the left side.

#x(log(4)+4log(5)-2log(10))=3log(10)-3log(5)#

#7#. Solve for #x#.

#x=(3log(10)-3log(5))/(log(4)+4log(5)-2log(10))#

#color(green)(|bar(ul(color(white)(a/a)x~~0.65color(white)(a/a)|)))#

Apr 16, 2016

#0.646#

Explanation:

#4^x*5^(4x+3)=10^(2x+3)#

#=>4^x*5^(4x)*5^3=10^(2x)*10^3#

#=>(4*5^4)^x*5^3=(10^2)^x*10^3#

Dividing both sides by #5^3*(10^2)^x#

#=>((4*5^4)/10^2)^x=10^3/5^3#

#=>((cancel4*cancel5xxcancel5xx5^2)/(cancel10xxcancel10))^x=1000/125=8=2^3#

Taking# log_10# on both sides and using property #logm^n=nlogm# we get
#=>log_10 5^(2x)=log_10 2^3#
#=>2x*log_10 5=3*log_10 2#

Dividing both sides by #(2*log_10 5)#

#=>x=(3*log_10 2)/(2*log_10 5)=0.646#