Question

Write each logarithmic equation as a single logarithm
(2logbx/3+3logby/4)-5logbz

Answer 1

please put some brackets for simplification. this seems very obsolete

Answer 2

(2/4(log(b)x) + 3/4(log(b)y)- 5log(b)z = (log(b)x^(2/4) +log(b) y^(3/4)) -log(b)z^5

Answer 3

can you explain how to do it

Answer 4

\[\log_{b}((x^{1/2} * y^{3/4})/z^5) \]
this is the correct way.

Answer 5

how did you get that

Answer 6

okay. i'll show you

Answer 7

first of all, what you gave me:
\[(1/2(\log_{b} x) +3/4(\log_{b} y) ) - 5\log_{b}z \]
correct me if m wrong

Answer 8

2/3(logbx)

Answer 9

(2/3) ! stupid of me overlooking that
in that case here is how it should look
\[((2/3) \log_{b}x + (3/4) \log_{b}y)-5\log_{b}z \]

Answer 10

which is equal to :
\[(\log_bx^{2/3} + \log_by^{3/4})-\log_bz^5\]

Answer 11

& since their base is common, u can group them like this using the basic rules of logarithm :
\[\log_b((x^{2/3} * y^{3/4})/z^5)\]

Answer 12

\[(\frac{ 2\log_{b}x }{ 3}+\frac{ 3\log_{b} y }{ 4})-5\log_{b} z\]

Answer 13

yes, i have given the evaluated answer to this equation :)

Answer 14

how do i do that

Answer 15

first of all remember that the (2/3) that is multiplied by \[\log_{b} x\] can also be written as \[\log_{b} x^{2/3}\] ie: we are just going in the opposite way of how logarithms are usually provided. for example:
\[\log_ba^2 = 2\log_ba\]

Answer 16

u know this rule right? it is possible to take the power of 'a' in the equation that i mentioned above & multiply it to the log.