Solve 4^(2x) = 7^(x−1)
how do you start this problem?

Question

Solve 4^(2x) = 7^(x−1)
how do you start this problem?

photon336 · Answer

$$4^{2x} = 7^{x-1}$$ right?

anonymous · Answer

yes thats it

photon336 · Answer

You can use logs to solve this

anonymous · Answer

oh okay, ive missed a few days of school though aand im behind how would i enter in into my calculator?

anonymous · Answer

a.2.35389
b.−2.35389
c.−1
d.1
these are the final answer options i have

photon336 · Answer

okay so see $$4^{2x} = 2x*\log(4)$$

anonymous · Answer

okay so would that me the other side would equal $$7^{x-1}=x-1*\log_{7} $$

anonymous · Answer

?

photon336 · Answer

so on the other side it becomes 
 
Yep :)

$$2xlog(4) = \log(7)(x-1)$$

photon336 · Answer

$$2x*\log(4) = x*\log(7) - \log(7)$$

anonymous · Answer

then you enter the whole equation into the calculator?

photon336 · Answer

well before that we've got to get x by itself

anonymous · Answer

oh okay, so to do that we take away one x from the right side?

photon336 · Answer

$$2x*\log(4)-x \log(7)  = -\log(7)$$

factor out a x

$$x(2\log(4)-\log(7) = -\log(7)$$

then divide that whole thing in the (  ) 

$$\frac{ -\log(7) }{ (2\log(4)-\log(7)) } = x $$

anonymous · Answer

oh okay so when you take the x away from the right side you have to take the log and everything with it then simplify it

anonymous · Answer

so then that would be what we enter into the calculator because x is by itself correct?

photon336 · Answer

yeah that's the simplified form you can enter that into a calculator and get the exact value of x

anonymous · Answer

okay so i got -2.35388 which issss

anonymous · Answer

b!? right?

photon336 · Answer

we can easily check this answer by plugging it into the original equation

anonymous · Answer

alrighty

anonymous · Answer

that worked for me :) ty so much!

photon336 · Answer

np anytime :)