Thread: Solving a (Tough) Exponential Function

So, please don't ignore me cause I'm only going to give you the equation but I've tried several different routes of attacking this equation and I can't seem to figure it out!

4^x + 6(4^(-x)) = 5

Hello,

Originally Posted by mike_302

So, please don't ignore me cause I'm only going to give you the equation but I've tried several different routes of attacking this equation and I can't seem to figure it out!

4^x + 6(4^(-x)) = 5

Multiplying both sides by $\displaystyle 4^x$ we get $\displaystyle 4^{2x}+6=5\times 4^x$ that is to say $\displaystyle \left(4^x\right)^2-5\times4^x+6=0$. Can you solve this equation ?

Yeeeeeaaaa, I can see how that works.... thanks a bunch! I was aiming for that quadratic form a couple of times, but I kept seeing the whole 4^2x as... simply that, 4^2x ... I didn't think of factoring it out to get "squared"

Thanks!

Hello, Mike!

Solve for $\displaystyle x\!:\;\;4^x + 6\cdot4^{-x} \:=\: 5$

Multiply by $\displaystyle 4^x\!:\;\;4^{2x} + 6 \:=\:5\cdot4^x\quad\Rightarrow\quad 4^{2x} - 5\cdot 4^x + 6 \:=\:0$

Factor: .$\displaystyle \left(4^x-2\right)\left(4^x - 3\right) \:=\:0$


Then we have:

. . $\displaystyle 4^x - 2 \:=\:0 \quad\Rightarrow\quad 4^x \:=\:2 \quad\Rightarrow\quad x \:=\:\log_4(2) \quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{2}}$

. . $\displaystyle 4^x - 3 \:=\:0\quad\Rightarrow\quad 4^x \:=\:3 \quad\Rightarrow\quad x \:=\:\log_4(3)\quad\Rightarrow\quad\boxed{ x \:=\:\frac{\ln(3)}{\ln(4)}} $

Thanks both of you! I shall remember this little..... trickery of 2x and x^2 (as exponents)

Let me know if I'm making a mistae by just reposting in this thread.... It's the exact same question, so I didn't think it was worthy of a thread wit hthe same name.

Here's the question:

log(3x-5)base4 = log(11)base4 + log(2)base4

I got it to

log(3x-5)base4 = log(22)base4

AHH! I figured it out.... Same bases. Nevermind :P

22=3x-5