How to Solve a Complex Algebraic Equation with Logs and Square Roots?

[Matrix]

Hi, I am having trouble finding x for the following exquation:
2^x=16^-3/4 * sqrt(2)

Any help on solving it would be much apprecitated.

 

[dduardo]

why don't you use logs.

2^x = 16^(-3/4) * sqrt(2)

x*ln(2) = ln(16^(-3/4) * sqrt(2) )

x = ln(16^(-3/4) * sqrt(2) ) / ln(2)

 

[jcsd]

reduce it further to:

x = -2.5

 

[dduardo]

jcsd, your i see you changed your reduction. The previous one was obviously wrong.

 

[Matrix]

Is it possible to find x not using logs?

 

[jcsd]

Duardo: No, it was perfectly correct.

 

[dduardo]

Originally posted by Matrix
Is it possible to find x not using logs?

yes.

you can convert 16 to a base 2

16^(-3/4) = (2^4)^(-3/4) = 2^(-3)

therefore

2^x = 2^(-3) * 2^(1/2)

2^x = 2^(-5/2)

x = -2.5

 

[Chrono]

This is probably off topic, but don't you just love logs?