MHB What is the Solution for $5^x+2=12$?

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The equation $5^x + 2 = 12$ simplifies to $5^x = 10$, leading to the solution $x = \frac{\ln(10)}{\ln(5)} \approx 1.4307$. A suggestion was made to use logarithm base 5 for clarity, but natural logarithms are more practical for most calculators. There was confusion regarding the original problem statement, as one participant initially interpreted it as $5^x + 2 = 126$. The thread was clarified when the title was updated to reflect the correct equation. The final solution for the original problem remains approximately 1.4307.
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Solve $5^x+2=12$
$5^x=10\implies x\ln(5)=\ln(10)\implies x=\dfrac{\ln(10)}{\ln(5)}\approx \boxed{1.4307}$

ok i think this is ok but typos or better steps maybe
 
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karush said:
Solve $5^x+2=12$
$5^x=10\implies x\ln(5)=\ln(10)\implies x=\dfrac{\ln(10)}{\ln(5)}\approx \boxed{1.4307}$

ok i think this is ok but typos or better steps maybe
Looks good. The only possible improvement I might suggest is to use [math]log_5[/math] instead of ln (then [math]x = log_5(10)[/math]) but in practical terms most calculators don't have a key to evaluate [math]log_5(10)[/math] so I'd use ln, too.

-Dan
 
You solved $5^x+ 2= 12$ but the title of this thread was "Solve $5^x+ 2= 126$". Which is it?
 
If the problem is actually $5^x+ 2= 126$ then $5^x= 124$, $log(5^x)= x log(5)= log(124)$ so $x= \frac{log(124)}{log(5)}$. Since $5^3= 125$ x will be a little less than 3. Using a calculator, x is about 2.99500933 .
 
County Boy said:
You solved $5^x+ 2= 12$ but the title of this thread was "Solve $5^x+ 2= 126$". Which is it?
sorry I edited the title to what I solved
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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