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How do you solve #(2-x)log4=(5-x)log2#?

Answer 1

Sadie Alexander

The answer is #x=-1#

log(4)=#log(2^2)#=#2log(2)#

So, you can simplify #log(2)# You have the linear equation #4-2x=5-x => x=-1#

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Answer 2

Elijah Allen

To solve ( (2 - x)\log_4 = (5 - x)\log_2 ):

Apply the property of logarithms: ( \log_a(b) = \frac{\log_c(b)}{\log_c(a)} ).
Rewrite the equation as ( \frac{\log(2 - x)}{\log(4)} = \frac{\log(5 - x)}{\log(2)} ).
Cross-multiply to eliminate the denominators: ( \log(2 - x) \cdot \log(2) = \log(5 - x) \cdot \log(4) ).
Apply the property ( \log_a(b) = \log_a(c) ) if and only if ( b = c ).
Solve the resulting equation ( \log(2 - x) \cdot \log(2) = \log(5 - x) \cdot \log(4) ) for ( x ).

Therefore, by solving the equation resulting from step 5, you can find the value(s) of ( x ) that satisfy the original equation ( (2 - x)\log_4 = (5 - x)\log_2 ).

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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.