Math Help - Logarithms :(

1. Logarithms :(

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

2. 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

3. Originally Posted by icecreamfrk09
3.log5 (1/25) =?
$\log_5(\frac{1}{25})$

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

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

= $-2\times 1$

= $-2$

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

5. 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.

6. Originally Posted by icecreamfrk09

4. logx = -0.7
This question really depends on the logs base, lets try base 10.

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

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

$x = 10^{-0.7}$

7. k thank you tkhunny i understand now

8. 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

9. 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 $b^{c}\;=\;a$