Logarthmic

[kpx001]

Express as a single logarthim
b = base
5log7t
-------------
3


answer = log7t^(5/3)


why would it be that? a # is - if it in denominator brought to numerator?[/list]




and

solve and check

7e^x-32=45
7e^x=77
xln7e = ln 77

ln77
-----
ln7e

x = 1.4745


what did i do wrong in this problem?

 

[stapel]

Because one of the log rules you learned said that anything multiplied onto a log can be turned into a power inside.

For instance, (3/7)log₂(5) = log₂(5^3/7).

Eliz.

 

[kpx001]

i understand the first part but i reditted and added a 2nd problem thing sry
if possible can u explain that one?

 

[jolly]

i understand the first part but i reditted and added a 2nd problem thing sry
if possible can u explain that one?

huh? :?

 

[soroban]

Hello, kpx001!

Solve and check: \(\displaystyle 7e^x\,-\,32\;=\;45\)

Add 32 to both sides: \(\displaystyle \;7e^x\;=\;77\)

Divide by 7: \(\displaystyle \;e^x\;=\;11\)

Take logs: \(\displaystyle \;\ln(e^x)\;=\;\ln(11)\)

Then: \(\displaystyle \;x\cdot\ln(e)\;=\;\ln(11)\)

Since \(\displaystyle \ln(e)\,=\,1\), we have: \(\displaystyle \,x\;=\;\ln(11)\)

 

[stapel]

i reditted and added a 2nd problem thing sry if possible can u explain that one?

huh?

Yeah, the cryptic kiddie-lingo is hard to follow, but the point is that the poster did a "stealth edit" to his post, adding a second question after I'd replied to the first question, and now he wants the answer.

Eliz.