Using Log to Undo an Exponent

[Jason76]

Let's see if this reasoning is correct:

\(\displaystyle 3^{x} = 9\)

\(\displaystyle log_3(3^{x})= log_3(9)\) - On the left side the \(\displaystyle log\) and exponent undo each other.

\(\displaystyle x = log_3(9)\) Use calculator to see that the \(\displaystyle log\) of \(\displaystyle 9 = 3\)

\(\displaystyle x = 3\)

 

[JeffM]

Let's see if this reasoning is correct:

\(\displaystyle 3^{x} = 9\)

\(\displaystyle log_3(3^{x})= log_3(9)\) - On the left side the \(\displaystyle log\) and exponent undo each other.

\(\displaystyle x = log_3(9)\) Use calculator to see that the \(\displaystyle log\) of \(\displaystyle 9 = 3\)

\(\displaystyle x = 3\)

As denis said SLOW DOWN

\(\displaystyle 3^x = 9 = 3^2 \implies 3^x = 3^2 \implies x = 2.\)

 

[Jason76]

WHOA! Slow down...can you not see that 3^3 = 27 ?

You don't have to use the \(\displaystyle log\) thing in this case, but in another case you might. At least it's good to the mechanics of it all.

 

[JeffM]

You don't have to use the \(\displaystyle log\) thing in this case, but in another case you might. At least it's good to the mechanics of it all.

Jason

Then get the mechanics right.

\(\displaystyle 3^x = 9 = 3^2 \implies \log_3(3^x) = \log_3(3^2) \implies x * log_3(3) = 2 * log_3(3) \implies x * 1 = 2 * 1 \implies x = 2.\)

 

[Jason76]

Jason

Then get the mechanics right.

\(\displaystyle 3^x = 9 = 3^2 \implies \log_3(3^x) = \log_3(3^2) \implies x * log_3(3) = 2 * log_3(3) \implies x * 1 = 2 * 1 \implies x = 2.\)

I see the error now.

\(\displaystyle 3^{x} = 9\)

\(\displaystyle log_3(3^{x}) = log_3(9)\)

\(\displaystyle x = 2\)