Teslim.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hello.

$$\text{If:}\;\;\;\;2^a=81\;\;\;\text{and}\;\;\;\;2^b=3\;\;\;\;\;\;\text{Find:}\;\frac{a}{b}$

$$A)\;9$

$$B)\;1$

$$C)\;2$

$$D)\;4$

I have tried nothing, because I don't know what to try!

$$\text{I used computer but like always:}$

$2^a=81

$$\text{log}\;(2^a)=\text{log}\;(81)$

$$a\;\text{log}\;(2)=\text{log}\;(81)$

$$a=\frac{\text{log}\;(81)}{\text{log}\;(2)}$

$$a=\text{log}_2(81)$

$$a=4\;\text{log}_2(3)$

$a\;\approx\;6.3398500028846$

And pretty much the same with the second one too:

$$b\;\approx\;1.5849625007212$

$$\text{therefore:}\;\frac{a}{b}\;\approx\;3.9999999999999$

But I would really like to know how to solve it without calculator!

Shadow · **Posted:** Fri, 13 Jan 2012 20:48:44 UTC

kreshnik wrote:

Hello.

$$\text{If:}\;\;\;\;2^a=81\;\;\;\text{and}\;\;\;\;2^b=3\;\;\;\;\;\;\text{Find:}\;\frac{a}{b}$

$$A)\;9$

$$B)\;1$

$$C)\;2$

$$D)\;4$

I have tried nothing, because I don't know what to try!

$$\text{I used computer but like always:}$

$2^a=81

$$\text{log}\;(2^a)=\text{log}\;(81)$

$$a\;\text{log}\;(2)=\text{log}\;(81)$

$$a=\frac{\text{log}\;(81)}{\text{log}\;(2)}$

$$a=\text{log}_2(81)$

$$a=4\;\text{log}_2(3)$

$a\;\approx\;6.3398500028846$

And pretty much the same with the second one too:

$$b\;\approx\;1.5849625007212$

$$\text{therefore:}\;\frac{a}{b}\;\approx\;3.9999999999999$

But I would really like to know how to solve it without calculator!

Taking logs of both sides in base 3, you get:

$a\log_3(2)=4$
$b\log_3(2)=1$

Now do division:

${a\log_3(2)\over b\log_3(2)}={4\over 1}$ . Now cancel like terms.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Shadow wrote:

Taking logs of both sides in base 3, you get:

$a\log_3(2)=4$
$b\log_3(2)=1$

Now do division:

${a\log_3(2)\over b\log_3(2)}={4\over 1}$ . Now cancel like terms.

Thank you very much.

Shadow · **Posted:** Fri, 13 Jan 2012 22:44:06 UTC

kreshnik wrote:

Shadow wrote:

Taking logs of both sides in base 3, you get:

$a\log_3(2)=4$
$b\log_3(2)=1$

Now do division:

${a\log_3(2)\over b\log_3(2)}={4\over 1}$ . Now cancel like terms.

Thank you very much.

Ooh! Alternatively, you can go from what you did and use the change of base formula:

$\log_c(d)={\log d\over\log c}$

Your work shows $a={\log 81\over\log 2}$ and $b={\log 3\over\log 2}$ , then you get ${a\over b}={\log 81\over\log 3}=\log_3(81)$

and one last way you could do it from the way you started is to use the log property $\log(a^x)=x\log a$ so that you can avoid the change of base formula and just get $${a\over b}={\log(3^4)\over\log 3}={4\log 3\over\log 3}=4$

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