(-2)^x = - 1/128 solve for x

does x=-7?

yes

Your answer is correct.

( - 2 )ˣ = - 1 / 128

Multiply both sides by 128

128∙( - 2 )ˣ = - 1

Raise both sides to the power 2

128²∙[ ( - 2 )² ]ˣ = ( - 1 )²

128 = 2⁷ , ( - 2 )² = ( 2 )² , so:

(2⁷)²∙(2²)ˣ = 1

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Remark:

(2⁷)² is the same (2²)⁷
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(2²)⁷∙(2²)ˣ = 1

4⁷∙4ˣ = 1

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Remark:

nᵃ∙ nᵇ = nᵃ⁺ᵇ

Any number raised to the powe zero is one.

So:

4⁷ ∙ 4ˣ = 4⁷⁺ˣ

4⁰ = 1
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4⁷⁺ˣ = 4⁰

Take the logarithms base 4 of both sides

log₄( 4⁷⁺ˣ ) = log₄( 4⁰ )

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Remark:

log aᵐ = m ∙ log a

log₄( 4 ) = 1

So:

log₄( 4⁷⁺ˣ ) = ( 7 + x ) ∙ log₄ ( 4 ) = ( 7 + x ) ∙ 1 = 7 + x

log₄= 4⁰ = 0 ∙ log₄( 4 ) = 0
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7 + x = 0

Subtract 7 to both sides

7 + x - 7 = 0 - 7

x = - 7

whoa! Can you say overkill??

Steve

solve for x mean SOLVE for x

Just say:

x = - 7 is not enough.

I showed the way to the solution.

To solve the equation (-2)^x = -1/128 for x, we can use logarithms.

Step 1: Take the logarithm of both sides of the equation. You can choose any base, but it is commonly done with base 10 or base e (natural logarithm). Let's use base 10 logarithm (log10) for this example:
log((-2)^x) = log(-1/128)

Step 2: Apply the logarithm property log(xy) = y log(x) to bring down the exponent x:
x log(-2) = log(-1/128)

Step 3: Evaluate the logarithms on the right side of the equation:
x log(-2) = log(1/128) [Using the property log(1/x) = -log(x)]

Step 4: Evaluate the logarithm on the left side of the equation. Note that logarithms of negative numbers are not defined in the real number system.
Since (-2) raised to any power does not give a real number, the equation has no real solutions in this case.