MHB -6.1.12 solve for x 3^{2x}=105

[karush]

solve for $\ x \quad 3^{2x}=105$
rewrite $x\ln3^2=x\ln{9}=\ln{105}$
hence $x=\dfrac{\ln{105}}{\ln{9}}$

can this be reduced more except for decimal

 

[HOI]

I would have done this a little differently:
2x ln(3)= ln(105)

x= ln(105)/2ln(3).

Of course, since 9= 3^2, ln(9)= 2ln(3) so that is the same as your answer. Whether it is "simpler" or not is a matter of personal choice.

 

[Prove It]

Why not just $\displaystyle x = \frac{1}{2}\log_3{\left( 105 \right) }$?

 

[karush]

change of base then..

 

[Prove It]

change of base then..

I'm asking, why are you even bothering to change the base at all? The base of your exponential equation is 3, surely the best base for your logarithm would therefore also be 3...

 

[HOI]

But my calculator doesn't HAVE a "logarithm base 3" key!

 

[Prove It]

But my calculator doesn't HAVE a "logarithm base 3" key!

And why are you bothering to use a calculator? Surely they would want an exact answer...