Logarithm inequality $ \log_{1/2} (A) > 3 $

Question

If $ \log_{1/2} (A) > 3 $ then how can I write it in exponential form?

And how to solve this?

score 4 · Accepted Answer · answered Dec 04 '17 at 07:50

4

$$3<\dfrac{\log_2A}{-1}\iff-\log_2A>3\iff\log_2A<-3\iff A<2^{-3}$$

answered Dec 04 '17 at 07:50

lab bhattacharjee

274,582

what if the logarithm's base were $ \frac{3}{2} $ instead of $ \frac{1}{2} $ – Maanav Roy Dec 04 '17 at 08:01
1

@MaanavRoy $(3/2)^x$ would be an increasing function of $x$. So, we would have $A >(3/2)^3$. – Dec 04 '17 at 08:08

score 2 · Answer 2 · answered Dec 04 '17 at 07:49

2

Note that $(1/2)^x$ is a (montonically) decreasing function of $x$. Thus since $\log_{1/2}(A) > 3$, $(1/2)^{\log_{1/2}(A)} = A < (1/2)^3 = 1/8$.

answered Dec 04 '17 at 07:49

eepperly16

7,232

score 1 · Answer 3 · answered Dec 04 '17 at 07:49

Consider the function $f(A) = (\frac{1}{2})^A$. Notice that it is strictly decreasing on its entire domain, so if $x < y$ then $f(x) > f(y)$. Thus, since $3 < \log_{1/2}(A)$, we know that: $$ \left( \frac{1}{2} \right)^3 > \left( \frac{1}{2} \right)^{\log_{1/2}(A)} = A $$ so that $A < 1/8.$

Logarithm inequality $ \log_{1/2} (A) > 3 $

3 Answers3