Solving the Equation: Two Answers?

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SUMMARY

The discussion centers on solving the equation \( 7.3^{2x + 1} = 10 \) using two methods. The first method is deemed incorrect due to the omission of \( \ln(7.3) \). The second method, which involves taking the logarithm base 10 of both sides, leads to the correct formulation \( x = \frac{1}{2}\left(\frac{\ln(10)}{\ln(7.3)}-1\right) \approx 0.0792 \). The rounded answer from this method is noted as inaccurate, indicating the need for precise calculations.

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siyanor
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I have an equation which answered it in two ways,which one is correct ?View attachment 163
 

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siyanor said:
I have an equation which answered it in two ways,which one is correct ?
https://www.physicsforums.com/attachments/163

The first method is not correct because ln(7.3) has simply disappeared.

The second method is correct but the rounded answer is not correct. $ x = \frac{1}{2}\left(\frac{\ln(10)}{\ln(7.3)}-1\right) \approx 0.0792 $
 
siyanor said:
I have an equation which answered it in two ways,which one is correct ?https://www.physicsforums.com/attachments/163

7.3^{2x + 1} \ = \ 10

siyanor,

(regarding the correct process of method #2),
because you have 10 on one side at that point,
that lends itself to taking log (base 10) of each
side:

\log_{10}7.3^{2x + 1} \ =\log_{10}10(2x + 1)\log(7.3) \ = \ 12x + 1 \ = \ \dfrac{1}{log(7.3)}And continue . . .
 

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