Logarithms :(

**icecreamfrk09** · July 20th 2009, 08:09 PM

I need help with these just cant figure them out

1. 3^(?) =1

2. 3^(?) = 1/3

3.log5 (1/25) =?

4. logx = -0.7

**pickslides** · July 20th 2009, 08:17 PM

Originally Posted by icecreamfrk09

1. 3^(?) =1

$3^x =1$

$\log_3(3^x) =\log_3(1)$

$x =\log_3(1)$

You can also consider $a^0=1$

Originally Posted by icecreamfrk09

I
2. 3^(?) = 1/3

this can be the same way as Q1

**pickslides** · July 20th 2009, 08:27 PM

Originally Posted by icecreamfrk09

3.log5 (1/25) =?

$\log_5(\frac{1}{25})$

= $\log_5(5^{-2})$

= $-2\times \log_5(5)$

= $-2\times 1$

= $-2$

**icecreamfrk09** · July 20th 2009, 08:28 PM

thank you both...so both one and two are the same answer??

**TKHunny** · July 20th 2009, 08:32 PM

What? How did you draw that conclusion? These are the most fundamental properties. You must know these or you WILL fail your exams.

$3^{1} = 3$
$5^{1} = 5$

$4^{0} = 1$
$6^{0} = 1$

$2^{-1} = 1/2$
$7^{-1} = 1/7$

Look at these until you no longer feel an urge to ask the question about 1 and 2 being the same.

**pickslides** · July 20th 2009, 08:36 PM

Originally Posted by icecreamfrk09

4. logx = -0.7

This question really depends on the logs base, lets try base 10.

$log_{10}(x) = -0.7$

$10^{log_{10}(x)} = 10^{-0.7}$

$x = 10^{-0.7}$

**icecreamfrk09** · July 20th 2009, 08:46 PM

k thank you tkhunny i understand now

**yeongil** · July 20th 2009, 08:48 PM

Originally Posted by TKHunny

What? How did you draw that conclusion? These are the most fundamental properties. You must know these or you WILL fail your exams.

To expand on this,

$3^{1} = 3$ because $\log_3 3 = 1$
$5^{1} = 5$ because $\log_5 5 = 1$

$4^{0} = 1$ because $\log_4 1 = 0$
$6^{0} = 1$ because $\log_6 1 = 0$

$2^{-1} = 1/2$ because $\log_2 \left(\frac{1}{2}\right) = -1$
$7^{-1} = 1/7$ because $\log_7 \left(\frac{1}{7}\right) = -1$

01

**TKHunny** · July 21st 2009, 08:33 PM

Originally Posted by icecreamfrk09

k thank you tkhunny i understand now

Get used to these two things and you will do well.

1) A Logarithm IS an Exponent.

2) These are equivalent statements, for appropriate a, b, and c

$log_{b}(a)\;=\;c$ and $b^{c}\;=\;a$

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