How do you solve this using logarithms?

Use logarithms to find #x#

#3(x)^0.6223=1.266#

Answer 1

#color(blue)(x=e^((ln(0.422))/0.6223)~~0.2499763052)#

#3x^(0.6223)=1.266#

Divide by 3:

#x^(0.6223)=0.422#

Taking natural logarithms of both sides:

#0.6223ln(x)=ln(0.422)#
Divide by #0.6223#:
#ln(x)=(ln(0.422))/0.6223#
Raising #bbe# to the power of each side:
#e^(ln(x))=e^((ln(0.422))/0.6223)#
#x=e^((ln(0.422))/0.6223)~~0.2499763052#
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Answer 2

To solve the inequality ( -8 < x - 11 < -6 ) using logarithms, we can follow these steps:

  1. Start with the original inequality: ( -8 < x - 11 < -6 ).

  2. Add 11 to all parts of the inequality to isolate (x): [ -8 + 11 < x - 11 + 11 < -6 + 11 ] This simplifies to: [ 3 < x < 5 ]

  3. Now, let's express the inequality using logarithms. We'll take the logarithm of each part of the inequality, using the natural logarithm (ln) for simplicity: [ \ln(3) < \ln(x) < \ln(5) ]

  4. Finally, simplify the expression: [ \ln(3) < \ln(x) < \ln(5) ]

So, the solution to the inequality using logarithms is ( \ln(3) < \ln(x) < \ln(5) ).

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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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