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# Can you solve for x (exponential function) watch

1. 3^x=11^2x+4

2. (Original post by cashmoneyorg)
3^x=11^2x+4

take logs of both sides?
3. it could be ln it could be log base anything. a log rule would be useful here, log-base-a B^c= c log-base-a B
4. Do you know that a^2x can be written as (a^x)^2?
5. Is the RHS 11^(2x+4) or 11^(2x) + 4?
6. 11^(2x+4)
Is the RHS 11^(2x+4) or 11^(2x) + 4?
11^(2x+4)
8. well as was previously said you take the logs on both sides
x log (3)= (2x +4) log(11)
then you distribute
x log (3)=2xlog(11) +4 log(11)
then move the x to the other side, factor the x out and do the remaining algebra
x(log(3)-2log(11))=4 log(11)

x=4log(11)/(log(3)-2log(11))

that gives you the correct result
9. bump*
10. (Original post by cashmoneyorg)
bump*
The previous post gave you the solution!
11. (Original post by RichE)
The previous post gave you the solution!
sorry didn't see it when i bumped
12. (Original post by RichE)
The previous post gave you the solution!
sorry didn't see it when i bumped
13. (Original post by mel3)
well as was previously said you take the logs on both sides
x log (3)= (2x +4) log(11)
then you distribute
x log (3)=2xlog(11) +4 log(11)
then move the x to the other side, factor the x out and do the remaining algebra
x(log(3)-2log(11))=4 log(11)

x=4log(11)/(log(3)-2log(11))

that gives you the correct result
thanks repped
14. (Original post by cashmoneyorg)
3^x=11^2x+4

i worked out x as being equal to 4log(11)/log(3)-2log(11) did it in 2 min!

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