SOLUTION: Solve the equation 3x^log2base5 + 2^logxbase5 = 64

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Question 1135116: Solve the equation
3x^log2base5 + 2^logxbase5 = 64
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

3x%5E%28log%285%2C2%29%29+%2B+2%5E%28log%285%2Cx%29%29+=+64+

2%5E%28log%285%2Cx%29%29+=+64+-3x%5E%28log%285%2C2%29%29

apply log rule: if f%28x+%29=g+%28x+%29 then log%28a%2Cf%28x%29%29=log%28a%2Cg+%28x+%29+%29

log%285%2Cx%5E%28log%285%2C2+%29%29+%29=log%285%2C%2864-3x%5E%28log%285%2C2+%29%29%29+%29

When the logs have the same base we have:

x%5E%28log%285%2C2+%29%29+=64-3x%5E%28log%285%2C2+%29%29%29+

x%5E%28log%285%2C2+%29%29%2B3x%5E%28log%285%2C2+%29%29++=64

4x%5E%28log%285%2C2+%29%29++=64

x%5E%28log%285%2C2+%29%29++=64%2F4

x%5E%28log%285%2C2+%29%29++=16.........take log of both sides

log%28x%5E%28log%285%2C2+%29%29%29++=log%2816%29

log%285%2C2+%29%2Alog%28x%29++=log%282%5E4%29

log%28x%29++=4log%282%29%2Flog%285%2C2+%29

log%28x%29++=4log%282%29%2F%28log%282+%29%2Flog%285%29%29

log%28x%29++=4%2F%281%2Flog%285%29%29

log%28x%29++=4log%285%29

log%28x%29++=log%285%5E4%29

x++=5%5E4

x++=625