Really need help to crack this one

[needhelpforbread]

Hiya, can someone aid me to solve the following equation for X?
Cant seem to find where to start with all the logarithms..
4^x - 4^(x-1) = 3^(x+1) - 3^x

 

[Deleted member 4993]

Hiya, can someone aid me to solve the following equation for X?
Cant seem to find where to start with all the logarithms..
4^x - 4^(x-1) = 3^(x+1) - 3^x

Can you simplify the following:

4^x - 4^(x-1)

= \(\displaystyle 4^x \ - \frac{4^x}{4}\)

similarly simplify the RightHandSide and continue....

Please share your work if you have further question.

 

[needhelpforbread]

Can you please elaborate further on how to get to the solution for x? Thank you a lot!

 

[Deleted member 4993]

Can you please elaborate further on how to get to the solution for x? Thank you a lot!

Have you completed the tasks I had defined in response #2.

Please tell us exactly where you are stuck. Please share your work.

 

[BigBeachBanana]

Can you please elaborate further on how to get to the solution for x? Thank you a lot!

The goal is to isolate x by itself. Currently, it appears on both sides of the equation. Think about how you can achieve that.

 

[needhelpforbread]

I have now found out that
4^x-4^(x-1)=3^(x+1)-3^x can be simplified to
3* 4^(x-1)= 2*3^x but am still struggling to arrive at the solution. Am i to use ln of both sides now?

 

[Deleted member 4993]

3* 4^(x-1)= 2*3^x

Can you use "log" now - try it and show us.....

 

[pka]

I have now found out that
4^x-4^(x-1)=3^(x+1)-3^x can be simplified to
3* 4^(x-1)= 2*3^x but am still struggling to arrive at the solution. Am i to use ln of both sides now?

The above can be reduced to [imath]2^{2x-3}=3^{x-1}[/imath]
That means that [imath](2x-3)\log(2)=(x-1)\log(3)[/imath]
Can you move forward?
[imath][/imath][imath][/imath][imath][/imath]