# Solve for x?

• Dec 10th 2012, 07:44 PM
maiden129
Solve for x?
Solve for x:
16^3x-2=(1/32)^2x-16

How to start this?

• Dec 10th 2012, 07:57 PM
MarkFL
Re: Solve for x?
A good place to start would be to restate the problem using bracketing symbols to let us know what the exponents actually are.
• Dec 11th 2012, 07:31 AM
Soroban
Re: Solve for x?
Hello, maiden129!

Quote:

Solve for x: 16^3x-2=(1/32)^2x-16

If that equation is: .$\displaystyle 16^{3x} -2 \:=\:\left(\tfrac{1}{32}\right)^{2x} - 16$

we have: .$\displaystyle (2^4)^{3x} \:=\:(2^{-5})^{2x} - 14 \quad\Rightarrow\quad 2^{12x} \:=\:2^{-10x} - 14$

. . . . . . . . $\displaystyle 2^{12x} + 14 - 2^{-10x} \:=\:0$

Multiply by $\displaystyle 2^{10x}\!\!:\;\;2^{22x} + 14\cdot 2^{10x} + 1 \;=\;0$

Good luck!

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If the equation is: .$\displaystyle 16^{3x-2} \:=\:\left(\tfrac{1}{32}\right)^{2x-16}$

we have: .$\displaystyle (2^4)^{3x-2} \:=\:(2^{-5})^{2x-16} \quad\Rightarrow\quad 2^{12x-8} \:=\:2^{-10x+80}$

Therefore: .$\displaystyle 12x - 8 \:=\:-10x + 80 \quad\Rightarrow\quad 22x \:=\:88 \quad\Rightarrow\quad x \:=\:4$