1. ## other log problem

ok this my other log problem i having, i make attempt here
express as a single log and evaluate

2 log5 50-log520-2log510+log5100

first i make this =log 5 50x20x100/100
=log5 1000
=log 5 10^3
=10

i not sure if this is correct ???

Thanks you for any helping !!

other problem i had was log 5 40 to calculate i get 5.3219 do i use calculator correctly ??

2. 2 log5 50-log520-2log510 +log5100

Let me write "log to the base 5 of 50" as Log(5)[50], so,

2Log(5)[50] -Log(5)[20] -2Log(5)[10] +Log(5)[100]

= Log(5)[50^2] -Log(5)[20] -Log(5)[10^2] +Log(5)[100]
= Log(5)[2500] -Log(5)[20] -Log(5)[100] +Log(5)[100]
= Log(5)[2500] -Log(5)[20]
= Log(5)[2500/20]
= Log(5)[125] -------------------the single log
= Log(5)[5*5*5]
= Log(5)[5^3]
= 3Log(5)[5]
= 3*1

-----------------------------------------
I forgot the Log(5)[40].

My calculator cannot give directly antilogs that are not based on 10 or on "e", so I need to convert first the Log(5)[40] into based on 10 or based on "e".

Conversion:
Log(b)[a] = Log(c)[a] / Log(c)[b]

In base 10,
Log(5)[40]
= Log(10)[40] / Log(10)[5]
= 1.60206 / 0.69897
= 2.292

Or, in base "e" or natural log,
Log(5)[40]
= Ln[40] / Ln[5]
= 3.688879 / 1.609438
= 2.292 --------------same as in base 10.

3. ## confused

I am not sure which is right on the explanations could you please let me know

4. Use the properties of logarithms:
$\log_ax+\log_ay=\log_a(xy)$
$\displaystyle\log_ax-\log_ay=\log_a\frac{x}{y}$
$\log_ax^n=n\log_ax$

5. Hello, gregorio!

I will assume you know the basic properties of logs . . .

Express as a single log and evaluate:

. . $2\!\cdot\!\log_5(50) - \log_5(20) - 2\!\cdot\!\log_5(10) + \log_5(100)$

We have: . $\log_5\!\left(50^2\right) - \log_5(20) - \log_5\!\left(10^2\right) + \log_5(100)$

. . $= \;\log_5(2500) - \log_5(20)\, \underbrace{- \log_5(100) + \log_5(100)}_{\text{equals }0}$

. . $= \;\log_5\!\left(\frac{2500}{20}\right) \;=\;\log_5(125) \;=\;\log_5\!\left(5^3\right)\;=\;3\!\cdot\!\under brace{\log_5(5)}_{\text{equals }1} \;=\;3\cdot1 \;=\;\boxed{3}$

6. ## yes now i seeing this

i not see before the 2 same that cancel out !!! thank you, i finding sometimes i making sillly mistakes, more concentrated needed!!!