How do you solve #9^(8x)=13^(-x+7)#?

Answer 1

#x=7-8log(9)/log(13)#

#9^(8x)=13^(-x+7)#

You need to use logarithms:

#8x. log(9)=(-x+7)log(13)#
#8log(9)/log(13)=-x+7#
#x=7-8log(9)/log(13)#

It is a very ugly solution.

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Answer 2

To solve the equation 98x=13x+79^{8x} = 13^{-x+7}, we can take the natural logarithm (ln) of both sides to remove the exponents. This allows us to solve for the variable xx.

First, take the natural logarithm of both sides:

ln(98x)=ln(13x+7)\ln(9^{8x}) = \ln(13^{-x+7})

Using the properties of logarithms, we can bring down the exponents:

8xln(9)=(x+7)ln(13)8x \ln(9) = (-x+7) \ln(13)

Next, distribute the natural logarithms:

8xln(9)=xln(13)+7ln(13)8x \ln(9) = -x \ln(13) + 7 \ln(13)

Now, isolate the variable xx by bringing all terms containing xx to one side of the equation:

8xln(9)+xln(13)=7ln(13)8x \ln(9) + x \ln(13) = 7 \ln(13)

Factor out the common factor of xx:

x(8ln(9)+ln(13))=7ln(13)x(8 \ln(9) + \ln(13)) = 7 \ln(13)

Finally, solve for xx by dividing both sides by the coefficient of xx:

x=7ln(13)8ln(9)+ln(13)x = \frac{7 \ln(13)}{8 \ln(9) + \ln(13)}

This is the solution for the equation 98x=13x+79^{8x} = 13^{-x+7}.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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