# Math Help - Logarithms Questions helpppp~

1. ## Logarithms Questions helpppp~

Back with more Qs^^

without calculator
solve for x :

logx = 3loga - logb +logc the answer is a^3c/b

i canceled all the logs and got x =a^3-b+c than i got stuck

simplify:
4^log2 3

i got to 3=2X than i got stuck the answer is 9... 3/2 can't get 9..

whats the value of X :
logx 8=3/4

32=3^x stuck

how can a square root b/c^2 = loga+1/2logb-2logc ?( can't find the root key on the keyboard>.<)

ty

2. ## Re :

Originally Posted by hovermet
Back with more Qs^^

without calculator
solve for x :

logx = 3loga - logb +logc the answer is a^3c/b

i canceled all the logs and got x =a^3-b+c than i got stuck

simplify:
4^log2 3

i got to 3=2X than i got stuck the answer is 9... 3/2 can't get 9..

whats the value of X :
logx 8=3/4

32=3^x stuck

how can a square root b/c^2 = loga+1/2logb-2logc ?( can't find the root key on the keyboard>.<)

ty

(1) $lgx=3lga-lgb+lgc$ , just apply the laws of logarithms

$lgx=lg[(a^3)(1/b)(c)]$

$lgx=lg(\frac{ca^3}{b})$
then cancel logs from both sides .

(2) $4^{log_23}$

$=2^{2log_23}$

$=2^{log_29}$
Do you know that $a^{log_ax}=x$

(3) $log_x8=3/4$

Try to convert it to index form and solve for X .

3. tyyy

i didnt know that u can multiply the 3^2 and that formula >.<

for number 3) whats that index form? is it the exponential form?

x^3/4 =8
3/4logx=8
3logx=32

stuck again the answer is 16

4)
how can "A"root b/c^2 = loga+1/2logb-2logc?

got a few more

5)given log80 = a which of the following is an expression for log2 in terms of a