Let r=log9 19, s=log5 19, t=log9 5. Find log9 (19^(1/5))

[snakeyesxlaw]

Let r = log9 19, s = log5 19, t = log9 5.

Write the following expressions in terms of r, s, and/or t. The change of base formula may be helpful in finding some of these logarithms.

log_9 19^(1/5) =

log_9 95 =

log_5 19(9^(log_81 5)) =

 

[ilaggoodly]

95 = 19*5
81 = 9^2
apart from that, we would appreciate some work....maybe? some of these are rather straightforward once you know the rules...do you know the rules?

 

[wjm11]

Let r = log9 19, s = log5 19, t = log9 5.

Write the following expressions in terms of r, s, and/or t. The change of base formula may be helpful in finding some of these logarithms.

log_9 19^(1/5) =

log_9 95 =

log_5 19(9^(log_81 5)) =

Hints:

1) log_9 19^(1/5) = (1/5)* log_9 19

2) 95 = 5*19

3) (log_81 5) = (log_9 5)/(log_9 81)

Hope this helps. As ilaggoodly says, please show some work.

 

[snakeyesxlaw]

Let r = log9 19, s = log5 19, t = log9 5

okay, stuck on the last:


log_5 19 (9^(log_81 5)) =

(9^(log_81 5)) = 5^(1/2) = log_5 19 (5^(1/2))

= (5^(1/2)) * s

when applying the substituted letters and submitting them, get it wrong. any suggestions?

 

[ilaggoodly]

you have
\(\displaystyle \L \log_5{19(9^{\log_{81}{5}})}\)
first seperate

\(\displaystyle \L \log_5{19} + \log_5{(9^{\log_{81}{5}})\)

next use your exponent rule

\(\displaystyle \L \log_5{19} + (\log_{81}{5}) (\log_5{9})\)

now ...change of base

\(\displaystyle \L \log_5{19} + (\frac{\log_9{5}}{\log_9 81}) (\frac{\log_9{9}}{\log_9{5}})\)

the rest is simply replacing it with your known variables