if [log_x(1/2) 2]^2 = log_x 2
Find x
(a) 8 (b) 16 (c) 4 (d) 32
the whole log is being squared.. and x(1/2) = underoot x
partap95 wrote:if [log_x(1/2) 2]^2 = log_x 2
Find x
(a) 8 (b) 16 (c) 4 (d) 32
the whole log is being squared.. and x(1/2) = underoot x
stapel_eliz wrote:partap95 wrote:if [log_x(1/2) 2]^2 = log_x 2
Find x
(a) 8 (b) 16 (c) 4 (d) 32
the whole log is being squared.. and x(1/2) = underoot x
I've heard of a log having a "base", but I've never heard of an "underoot". Please reply with your book's definition of "underoot". Thank you!
partap95 wrote:The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2).
stapel_eliz wrote:partap95 wrote:The base of the first log is x(1/2). Ignore the underoot.(it just refers to the power 1/2).
Do you mean that the base of the logarithm on the left-hand side is "the square root of x"? Is "x" he base of the log on the right-hand side? So the equation is as follows?
. . . . .
When you reply, please show what you've tried so far, even if it was just plugging the various answer-options into the equation to see if they work. Thank you!
stapel_eliz wrote:So the equation is as follows?
. . . . .
partap95 wrote:Yes that's the equation. I just plugged in the answers with 16 being the correct answer. Is there not any other way to solve this?
little_dragon wrote:stapel_eliz wrote:So the equation is as follows?
. . . . .partap95 wrote:Yes that's the equation. I just plugged in the answers with 16 being the correct answer. Is there not any other way to solve this?
i think U could use base change
log_sqrt[x](2)=log_x(2) / log_x(sqrt[x])=log_x(2) / (1/2)log_x(x)
& log_x(x)=1 so
log_x(2)/(1/2)=2log_x(2)
sqrd is 4(log_x(2))
then eqn is 4(log_x(2))^2=log_x(2)
4(log_x(2))^2-log_x(2)=0
log_x(2) [4log_x(2)-1]=0
log_x(2)=0 => x=1 but base cant be 1
4log_x(2)-1=0 => log_x(2)=1/4
x^(1/4)=2 => x=2^4=16
plz rite back if that doesnt make sense
thnx
partap95 wrote:2log a = log b
log a^2 = log b
a^2 = b
Why cannot 2 log a = log b; 2a=b work?