How do you solve #4(7^(x + 2)) = 9^(2x - 3)#?

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Answer 1 · 2018-04-22T14:28:43

#x=(-3ln(9)-2ln(7)-ln(4))/(ln(7)-2ln(9))#

you have to log the equations

#4*7^(x+2)=9^(2x-3)#

Use either natural logs or normal logs #ln# or #log# and log both sides

#ln(4*7^(x+2))=ln(9^(2x-3))#

First use the log rule that states #loga*b=loga+logb#

#ln(4)+ln(7^(x+2))=ln(9^(2x-3))#

Remember the log rule that states #logx^4=4logx#

#ln(4)+(x+2)ln(7)=(2x-3)ln(9)#

#ln(4)+xln(7)+2ln(7)=2xln(9)-3ln(9)#

Bring all the #xln# terms to one side

#xln(7)-2xln(9)=-3ln(9)-2ln(7)-ln(4)#

Factorise the x out

#x(ln(7)-2ln(9))=(-3ln(9)-2ln(7)-ln(4))#

#x=(-3ln(9)-2ln(7)-ln(4))/(ln(7)-2ln(9))#

Solve on the calculator using the ln button or if your calculator doesn't have it use the log base 10 button.