(-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.