How to Solve Logarithmic Equations with Different Bases?

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To solve the equation log2x + log4x = 5, it's essential to clarify the interpretation of the logarithmic terms. If interpreted as log(2x) + log(4x) = 5, it simplifies to log(6x) = 5, making it straightforward to solve. However, if the intention is log2(x) + log4(x) = 5, converting log4(x) to a base of 2 is necessary, using the formula log_a(x) = log_b(x)/log_b(a). This conversion allows for solving the equation with a common base. Properly defining the logarithmic expressions is crucial for finding the correct solution.
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Solve the equation log2x + log4x= 5. To start, should I change this to an exponential... I am stuck because I have only done log questions that have the same base.
 
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Convert them to the same base log_a(x)=log_b(x)/log_b(a).
 
emma3001 said:
Solve the equation log2x + log4x= 5. To start, should I change this to an exponential... I am stuck because I have only done log questions that have the same base.
What do you mean by "log2x+ log4x= 5"? I would interpret that as log(2x)+ log(4x)= 5 so log(6x)= 5 which is easy. If you mean "log_2(x)+ log_4(x)= 5" then use Dick's hint.
log_4(x)= log_2(x)/log_2(4)= ?
 
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Dick said:
Convert them to the same base log_a(x)=log_b(x)/log_b(a).

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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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