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

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Answer 1 · 2016-07-10T00:46:24

#x = (4ln(5)-2ln(10))/(5ln(10)+ln(5))~~0.1397#

Using the property that #ln(a^x)=xln(a)#:

#10^(5x+2) = 5^(4-x)#

#=> ln(10^(5x+2)) = ln(5^(4-x))#

#=>(5x+2)ln(10) = (4-x)ln(5)#

#=>5ln(10)x+2ln(10) = 4ln(5)-ln(5)x#

#=>5ln(10)x+ln(5)x = 4ln(5)-2ln(10)#

#=>(5ln(10)+ln(5))x = 4ln(5)-2ln(10)#

#=>x = (4ln(5)-2ln(10))/(5ln(10)+ln(5))~~0.1397#