The value of $a^{\log_b x}$ where

Question

a=0.2

$b=\sqrt 5$

$x=\frac 14 + \frac 18 + \frac {1}{16}.....$

Since x is a GP, common ratio ‘r’ is $\frac 12$

Then $$x=\frac 12$$ So $$(0.2)^{log_{\sqrt 5}\frac 12}$$ I don’t know how to simply it further. Using a calculator isn’t allowed, and log tables aren’t given either. I don’t think it’s possible, but I wanted a second opinion nonetheless.

The answer is 4.

Thanks!

maybe you could rewrite $0.2$ as $\left(\sqrt{5}\right)^{-2}$ to proceed? — Unknown, Sep 11 '19 at 17:45

score 3 · Accepted Answer · answered Sep 11 '19 at 17:44

3

$$\log_\sqrt5(1/2)=\frac{\ln(1/2)}{\ln\sqrt5}=\frac{\ln(2^{-1})}{\ln(5^{1/2})}=\frac{-\ln2}{(1/2)\ln5}=-2\frac{\ln2}{\ln5}=-2\log_5 2$$

$$(0.2)^{\log_\sqrt5(1/2)}=(5^{-1})^{-2\log_5 2}=5^{2\log_5 2}=5^{\log_5(2^2)}=2^2=4$$

answered Sep 11 '19 at 17:44

mr_e_man

5,364

score 1 · Answer 2 · answered Sep 11 '19 at 19:33

1

Note that $a=\frac{1}{b^2}$. So, you get

$$a^{\log_b x}=\frac{1}{b^{2\log_b x}} = \frac{1}{x^2} = 4$$

answered Sep 11 '19 at 19:33

trancelocation

32,243

The value of $a^{\log_b x}$ where

2 Answers2