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# Logarithms watch

1. Solve the equation 5^2x - (16 x 5^x) + 63

Is this correct?

2x log 5 - xlog(16x5) +log 63

X(2log5-log(16x5))=-log 63
X=-log63/(2log5-log(16 x 5)

= 3.562
2. No.
Consider it as a quadratic equation.
3. (Original post by Custardcream000)
Solve the equation 5^2x - (16 x 5^x) + 63

Is this correct?

2x log 5 - xlog(16x5) +log 63

X(2log5-log(16x5))=-log 63
X=-log63/(2log5-log(16 x 5)

= 3.562
The original equation is not equal to anything... =0?

And you can't take the log of each individual term - log(1+2x) is not equal to log(1) + log(2x), which is equal to log(2x).
4. (Original post by SeanFM)
The original equation is not equal to anything... =0?

And you can't take the log of each individual term - log(1+2x) is not equal to log(1) + log(2x), which is equal to log(2x).
Yeah forgot to put =0

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