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How do you solve for x?: #log_2(2^(x+3))=15#?

Answer 1

Abigail Allen

#x=12#

#log_2(2^(x+3))=15=># if #log_a(x)=yhArrx=a^y# , so:

#2^(x+3)=2^15=>#when the same bases to different exponents are

equal, the exponents also have to be equal, so:

#x+3=15=># subtract 3 from both sides:

#x=12#

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Answer 2

Eva Adamson

To solve for ( x ) in the equation ( \log_2(2^{x+3}) = 15 ), you can use the property of logarithms that states ( \log_b(b^x) = x ). Apply this property to simplify the equation. Then solve for ( x ).

( \log_2(2^{x+3}) = 15 )

Since ( \log_b(b^x) = x ), we have:

( x + 3 = 15 )

Now, solve for ( x ):

( x = 15 - 3 )

( x = 12 )

So, the solution for ( x ) is ( x = 12 ).

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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.