Solve for x?

**maiden129** · December 10th 2012, 07:44 PM

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

How to start this?

Thanks in advance.

**MarkFL** · December 10th 2012, 07:57 PM

A good place to start would be to restate the problem using bracketing symbols to let us know what the exponents actually are.

**Soroban** · December 11th 2012, 07:31 AM

Hello, maiden129!

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

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

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

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

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

Good luck!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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

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

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

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