Finding X from Exponential Equations

Question

$$2^x \cdot 4^{1-x}= 8^{-x}$$

I wrote all the base numbers as a power of 2 but I'm not sure what to do after.

Answer 1

Using $2^x\cdot 4^{1-x}=8^{-x}$ from the comments, write everything as a power of two: $2^x\cdot (2^2)^{1-x}=(2^3)^{-x}$, and then use the rules of exponents. You need these: $(a^b)^c=a^{bc}$ and $a^ma^n=a^{m+n}$.

Once you have simplified the expression to something like $2^a=2^b$, you can take log base 2 to get $a=b$, and then solve for $x$.

Answer 2

If you are referring to $2^x * 4^{1-x} = 8^{-x}$ then here are the steps

$2^x * 4^{1-x} = 8^{-x}$

$2^x * 4^1 4^{-x} = 8^{-x}$

$2^x * 4\frac{1}{4^{x}} = \frac{1}{8^{x}}$

$4=\frac{4^x}{8^x 2^x}$

$4={\frac{1}{4}}^x$

$log_{\frac{1}{4}}(4)=x$

$x=-1$

Hope this helps!

Answer 3

$$2^x \cdot 4^{1-x}= 8^{-x}$$ $$\implies 2^x\cdot (2^2)^{1-x}=(2^3)^{-x}$$ $$\implies2^x\cdot 2^{2-2x}=2^{-3x}$$ $$\implies 2^{x+2-2x}=2^{-3x}$$ $$\implies 2^{2-x}=2^{-3x}$$ comparing both side we have, $$2-x=-3x$$ $$\implies -2x=2$$ $$\implies x=-1$$