One Problem involving Exponents

[Vertciel]

Hello everyone,

I am having trouble with the following problem and I would appreciate any help or hints.

Thank you!

----

1. If \(\displaystyle 10^{2y} = k\), find \(\displaystyle 10^{-y}\).

My Work :

\(\displaystyle log 10^{2y} = log k\)

\(\displaystyle 2y = \frac{log10}{log k}\)

\(\displaystyle y = \frac{log 10}{2 log k}\)

\(\displaystyle y = \frac{1}{2 log k}\)

Therefore:
\(\displaystyle 10^{-y} = \frac{1}{10^\frac{1}{2log k}}\)

However, the provided answer for this problem is:

\(\displaystyle 10^{-y} = \frac{1} \sqrt{k}\)

 

[stapel]

\(\displaystyle log 10^{2y} = log k\)

\(\displaystyle 2y = \frac{log10}{log k}\)

\(\displaystyle y = \frac{1}{2 log k}\)

You had:

. . . . .10[sup:15g4eg25]2y[/sup:15g4eg25] = k

You took the base-10 log of either side:

. . . . .log[sub:15g4eg25]10[/sub:15g4eg25](10[sup:15g4eg25]2y[/sup:15g4eg25]) = log[sub:15g4eg25]10[/sub:15g4eg25](k)

...and simplified:

. . . . .2y log[sub:15g4eg25]10[/sub:15g4eg25](10) = log[sub:15g4eg25]10[/sub:15g4eg25](k)

. . . . .2y = log[sub:15g4eg25]10[/sub:15g4eg25](k)

. . . . .y = log[sub:15g4eg25]10[/sub:15g4eg25](k) / 2

How did you end up with "y = log[sub:15g4eg25]10[/sub:15g4eg25](10) / [2 log[sub:15g4eg25]10[/sub:15g4eg25](k)]"? How did log[sub:15g4eg25]10[/sub:15g4eg25](k) end up in the denominator?

Eliz.

 

[pka]

You know that k must be positive. So start by using the square root.
\(\displaystyle 10^{2y} = k\quad \Rightarrow \quad 10^y = \sqrt k \quad \Rightarrow \quad y = \log \left( {\sqrt k } \right)\).
The rest is easy.

 

[Deleted member 4993]

1. If \(\displaystyle 10^{2y} = k\), find \(\displaystyle 10^{-y}\)....

the shorter way would be:

\(\displaystyle 10^{2y} = k\), find \(\displaystyle 10^{-y}\)

\(\displaystyle {[10^{y}]}^2 = k\)

\(\displaystyle \sqrt{{10^{y}}^2} = \sqrt{k}\),

\(\displaystyle 10^{y} = \sqrt{k}\)

\(\displaystyle \frac{1}{10^{y}} = \frac{1}{\sqrt{k}}\)

\(\displaystyle 10^{-y} = \frac{1}{\sqrt{k}}\)

Sometimes, it is important to find the shorter way to answer (like when taking SAT etc.).

Things to think about:

Why didn't I consider \(\displaystyle \pm\\) while taking \(\displaystyle \sqrt{}\\) of the expressions?

 

[soroban]

Hello, Vertciel!

We don't need logs for this problem . . .

\(\displaystyle 1.\;\text{ If }10^{2y}\: =\: k\text{, find }\:10^{-y}\).

\(\displaystyle \text{Raise both sides to the }\frac{1}{2}\text{ power: }\;\left(10^{2y}\right)^{\frac{1}{2}} \:=\:k^{\frac{1}{2}}\quad\Rightarrow\quad 10^y \:=\:\sqrt{k}\)

\(\displaystyle \text{Raise both sides to the -}1\text{ power: }\;\left(10^y\right)^{\text{-}1} \:=\:\left(\sqrt{k}\right)^{\text{-}1}\)

\(\displaystyle \text{Therefore: }\;10^{-y} \:=\:\frac{1}{\sqrt{k}}\)

 

[Vertciel]

Thank you to everyone for their replies. I particularly enjoyed the variety of ways to solve this problem.

@stapel: That was probably a careless algebraic mistake.