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mstehman88 · May 12, 2009, 11:32 AM

I need help on solving the 2 log equations I attached to this message, I may have got the second one done but need to check it.

1. is Log base 10 to the x = log base 3 to the 9

2. 9 to the 7-3x = 5

Please let me know how to work them both out by an image or step by step text.

Thank you!
Matt

Perito · May 12, 2009, 11:46 AM

Originally Posted by ;

is Log base 10 to the x = log base 3 to the 9

Is this what you mean? Log base 10 of x = log base 3 of 9?

$log_{10}(x) = log_3(9)$

The important identity to remember is this one:

$log_b(x) = \frac {log_k(x)}{log_k(b)$

so

$log_3(9) = \frac {log_{10}(9)}{log_{10}(3)}$

make that substitution and solve for x. Unknown008 noted that 9 is a whole-number log of 3. That's a better way to do the problem if you see that.

#2:
$9^{(7-3x)} = 5$

$log(9^{(7-3x)}) = log(5)$

$(7-3x)log(9) = log(5)$

$x = \frac { 7log(9) - log(5)}{3log(9)}$

Is this what you got?

Unknown008 · May 12, 2009, 11:46 AM

1.Ok, firstly,

$log_{10}x = log_{3}9$

$log_{10}x = log_{3}3^2$

$log_{10}x = 2log_{3}3$

$log_{10}x = 2(1)$

$x = 10^2$

$x = 100$

You should know your index - log relation;

$log_{a}b = c$

$b = a^c$

2. Use the formula gave to have

$9^{7-3x} = 5$

$7-3x = log_{9}5$

Type the value of log_9 5 in your calculator and solve for x

Did that help? :)

Unknown008 · May 12, 2009, 11:48 AM

LOL Perito! :p