Hello, Angel!
For example: .\(\displaystyle 2^x\,=\,16\;\;\Rightarrow\;\;x = 4\) . (I find this by 2*2*2*2)
Q1: Is there a better way to do this?
If I had \(\displaystyle 3^x\,=\,243\), it would take a while to get to \(\displaystyle x = 5\) without a calculator on a test.
Without a calculator, that's the only way to do it . . . sorry.
\(\displaystyle x^4\,=\,16\;\;\Rightarrow\;\;x = 2\) . (I find this by taking the 4th root of 16)
BUT ... When the answer isn't obvious, how do I find the value of x?
For example:
\(\displaystyle Q2:\; 8^x\,=\,0.25\)
I know \(\displaystyle x = -\frac{2}{3}\), but I don't know how to get that answer
\(\displaystyle Q3:\;10^x\,=\,\sqrt{10}\)
I know \(\displaystyle x = \frac{1}{2}\), but why?
\(\displaystyle Q4:\;5^x\,=\,0.2\)
Again, I know \(\displaystyle x = -1\), but don't understand why
Are you familiar with the "Equal bases, equal exponents" rule?
If we have:
.\(\displaystyle b^x\,=\,b^y\), then \(\displaystyle x\,=\,y\).
. . Given two equal exponentials, if the bases are equal, then the exponents are equal.
\(\displaystyle Q3:\;10^x\,=\,\sqrt{10}\)
. . . . Try to get the same base on both sides.
. . . . Hey, they <u>are</u> the same . . . both 10's.
. . . . So their exponents are equal:
. . . . . . \(\displaystyle 10^x\,=\,10^{\frac{1}{2}}\;\;\Rightarrow\;\;x\,=\,\frac{1}{2}\)
\(\displaystyle Q2:\;5^x\,=\,0.2\)
. . . . Try to get the same base on both sides.
. . . . The left side has base \(\displaystyle 5\).
. . . . The right side is: \(\displaystyle 0.2\,=\,\frac{1}{5}\,=\,5^{-1}\) . . . there!
. . . . We have:
.\(\displaystyle 5^x\,=\,5^{-1}\;\;\Rightarrow\;\;x\,=\,-1\)
\(\displaystyle Q1:\;8^x\,=\,0.25\)
. . . . Try to get the same base on both sides.
. . . . The left side has base \(\displaystyle 8\).
. . . . The right side is: \(\displaystyle 0.25\,=\,\frac{1}{4}\,=\,4^{-1}\)
.(base 4)
Oh-oh!
.We have two
different bases (8 and 4).
. . Now what?
Sometimes we have to change <u>both</u> bases . . .
The left side is:
.\(\displaystyle 8^x\,=\,(2^3)^x\,=\,2^{3x}\)
The right side is:
.\(\displaystyle 4^{-1}\,=\,(2^2)^{-1}\,=\,2^{-2}\)
. . So we have:
.\(\displaystyle 2^{3x}\,=\,2^{-2}\) . . . same base!
Equate exponents:
.\(\displaystyle 3x\,=\,-2\;\;\Rightarrow\;\;\,=\,-\frac{2}{3}\) . . .
There!