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

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?

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? :)

LOL Perito! :p