[MOVED] Logic Behind These Steps: log_16 (64) = X

[Lime]

The problem:

log_16 (64) = X ... Imagine the 16 as subscript

Solution from the answer book:

(16)^X = 64 ... I get this
X = 3/2 ... I don't get this

The rationale, I think, behind that is written as such:

(16)^1/2 = square root of 16 = 4 & 4^3 = 64) :?

Please help me on this.

 

[galactus]

Change of base formula:

\(\displaystyle \L\\\frac{log(64)}{log(16)}=\frac{3\sout{log(4)}}{2\sout{log(4)}}=\frac{3}{2}\)

 

[Lime]

How did you arrive at the point you divide log(64) by log(16)?

 

[pka]

Alternatively,
\(\displaystyle \L
\begin{array}{l}
\log _b (A) = B \Leftrightarrow b^B = A \\
\log _{16} (64) = X \Leftrightarrow 16^X = 64\quad \Rightarrow \quad X = \frac{3}{2} \\
\end{array}.\)

 

[stapel]

How did you arrive at the point you divide log(64) by log(16)?

By applying the "change of base" formula:

. . . . .log_p(q) = log_r(q)/log_r(p)

If this hasn't been covered in your class yet, then use what you know about the relationship between 16, 4, and 64, relating 16 and 64 to powers (possibly fractional) of 4.

Eliz.

 

[soroban]

Hello, Lime!

Solve: \(\displaystyle \,\log_{16}(64)\:=\:x\)

We have: \(\displaystyle \,\log_{16}(64)\:=\:x\;\;\Rightarrow\;\;16^x\:=\:64\)

Get the same base on both sides: .\(\displaystyle (4^2)^x\:=\:4^3\;\;\Rightarrow\;\;4^{2x}\:=\:4^3\)

Then we have: \(\displaystyle \,2x\:=\:3\;\;\Rightarrow\;\;x\:=\:\frac{3}{2}\)