MHB Solving the Equation: Two Answers?

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The discussion centers on solving the equation 7.3^(2x + 1) = 10, which yielded two different answers. The first method is deemed incorrect due to the omission of ln(7.3). The second method is correct, providing the solution x = 0.0792, although the rounded answer is noted as inaccurate. A suggestion is made to take the logarithm of both sides to clarify the solution process. The focus remains on ensuring the correct application of logarithmic principles to solve the equation accurately.
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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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