Logarithms !!!!!?

Question

who can answer this with all the steps ???

show that :

2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32 = 5 - 5 log_10 5

who can answer this with all the steps ???

show that :

2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32 = 5 - 5 log_10 5

For ***** SAKE THE QUESTION IS RIGHT !!!!

Answer 1

2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32 = 5 - 5log_10 5

LHS = 2/3 + (3/2)*1.2041 - (3/5)*1.5051
LHS = 2/3 + 1.8062 - 0.9031
LHS = 0.6667 + 1.8062 - 0.9031
LHS = 1.5698

RHS = 5 - 5*0.6990
RHS = 1.5698

Answer 2

Simplify the right side:
5 - 5 log_10 5 = 5 log_10 10 - 5 log_10 5 = 5(log_10 10 - log_10 5) = 5 log_10 2 = log_10 32

Simplify the left side:
2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32 =

2/3 + log_10 16^3/2 - log_10 32^3/5 =

2/3 + log_10 64 - log_10 8 = 2/3 - log_10 8

2/3 - log_10 8 is not equal to log_10 32

However:

2/3 log_10 8 + 3/2 log_10 16 - 3/5 log_10 32 =

2/3 (3 log_10 2) + 3/2 (4 log_10 2) - 3/5(5 log_10 2)

= (2 + 6 - 3 )log_10 2 = 5 log_10 2 = log_10 32

Answer 3

Havnt done this in a while but i'll give it a try:


2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32 = 5 - 5 log_10 5
2/3 + log_10 (64) - log_10 (8) = 5 - 5log_10 (5)
2/3 + log(64 / 8) = 5 - 5log_10 (5)
2/3 + log_10 (8) = 5 - 5log_10 (5)

Sorry buddy, Cant help you any further... anway good luck finding your answer

Answer 4

here are some basic properties of logs.

log_a (xy) = log_a(x) + log_a(y)
log_a(x/y) = lot_a(x) - log_a(y)
b*log_a(x) = log_a (x^b)
log_a(x) = ln(x) / ln(a)
so

2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32
=
(2/3)*ln(10) / ln(10) + (3/2)*ln(16)/ln(10) - 3/5 ln(32) / ln(10)

Multiply both sides (of your original equation) by ln(10) to get

(2/3)*ln(10)+ (3/2)*ln(16)- 3/5 ln(32) =
5*ln(10) - 5*ln(10) * ln(5) / ln(10)

Simplify both sides (the right using the above mentioned rules for logs).

ln(10^(2/3)* 64 / 8) = 5*ln(10) - 5*ln(32)

ln(8*10^(2/3)) = ln(2^5)
...
actually what you said is false, I just double checked it with a computer algebra system to be sure.


2/3 log_10 10 + 3/2 log_10 16 - 3/5 log_10 32 = 1.56976

5 - 5 log_10 5 = 1.50515

If you are claiming your question to be right then you must define what the symbols you are using mean because they ARE NOT universal. I very much doubt that the Mathematica program I used to check my answers made a mistake because it is professional software. This is what I'm reading your question as

2/3 * LogBase10(10) + 3/2*LogBase10(16) - 3/5 LogBase10 (32)
=
5 - 5*LogBase10(5)

And THAT is a false statement.

Answer 5

The question is wrong. 2/3 log 10 + 3/2 log 16 - 3/5 log 32 ≈ 1.5698, but 5 - 5 log 5 ≈ 1.5051. These two expressions cannot possibly be equal.

Answer 6

log10(10) = 1
log10(16) = 4 • log10(2)
log10(32) = 5 • log10(2)
log10(5) = log10(10 / 2) = log10(10) - log10(2) = 1 - log10(2)

2 • log10(10) / 3 + 3 • log10(16) / 2 - 3 • log10(32) / 5 = 5 - 5 • log10(5)
2 • [1] / 3 + 3 • [4 • log10(2)] / 2 - 3 • [5 • log10(2)] / 5 = 5 - 5 • [1 - log10(2)]
2 / 3 + 6 • log10(2) - 3 log10(2) = 5 - 5 + 5 • log10(2)
2 / 3 + 3 • log10(2) = 5 • log10(2)
2 / 3 = 2 • log10(2)
1 / 3 = log10(2)
implies ³√(10) = 2, or 2³ = 10.

Sorry, but this problem isn't a correct one... it simplifies to a false statement.

Answer 7

(2/3)log(10)10 + (3/2)log(10)16 - (3/5)log(10)32 = 5 - 5log(10)5

log(10)(10^(2/3)) + log(10)(16^(3/2)) - log(10)(32^(3/5)) = 5 - log(10)(5^5)

log(10)((10^(2/3) * 64)/8) = 5 - log(10)3125

log(10)(10^(2/3) * 8) = 5 - log(10)(3125)

(log(10^(2/3) * 8))/(log(10)) = 5 - (log(3125)/log(10))

log(10^(2/3) * 8) = 5 - log(3125)

log(10^(2/3) * 8) + log(3125) = 5

log(10^(2/3) * 8 * 3125) = 5

log(25000 * 10^(2/3)) = 5

25000 * 10^(2/3) = 10^5

25000 * 10^(2/3) = 100000

10^(2/3) = 4

Sorry, but they don't work out.

Answer 8

Using a caculator, the LHS works out to 0.90309 and the RHS works out to 1.5051. So the question is incorrect.

Answer 9

can you plz use brackets other wise there may be 19 differnt calculations to be made contact me at [e-mail]

Answer 10

hey they all told ur qn is rong n u told ur isnt
i will support you !!!!
[ u know what i want in my status :P ]