log (2 Viewers)

[jessi90]

simplify: 10^log₁₀³

 

[jessi90]

show working please!

also: log₁₀²⁵ / log₁₀⁵

 

[3unitz]

log₁₀25
log₁₀5

Spoiler

= log25 - log5
= 0.69897000433601880478626110527601
= 0.699 (3dp)

this is wrong

answer:

log 25 / log 5
= log (5^2) / log 5
= 2 log 5 / log 5
= 2

 

[vds700]

isn't the first one 10 ^{log (base 10) 3}

let y = 10^{log (base 10) 3}
take the base 10 log of both sides
log₁₀ y = log₁₀ ^{log (base 10) 3}
log ₁₀ y = log₁₀3 . log₁₀10
log₁₀y = log₁₀ 3

therefore y = 3

 

[ticky2002]

this is wrong

answer:

log 25 / log 5
= log (5^2) / log 5
= 2 log 5 / log 5
= 2

I thought when it was division you minus the answers..?

 

[independantz]

I thought when it was division you minus the answers..?

log(a/b)=loga-logb

not,loga/logb=loga-logb

 

[ticky2002]

log(a/b)=loga-logb

not,loga/logb=loga-logb

Oh okay, thanks for clearing that up.

 

[jessi90]

isn't the first one 10 ^{log (base 10) 3}

let y = 10^{log (base 10) 3}
take the base 10 log of both sides
log₁₀ y = log₁₀ ^{log (base 10) 3}
log ₁₀ y = log₁₀3 . log₁₀10
log₁₀y = log₁₀ 3

therefore y = 3

sorry how can u do the line in bold? is there a specific rule or what? ty

 

[3unitz]

sorry how can u do the line in bold? is there a specific rule or what? ty

let, y = 10^(log 3)

log y = log [10^(log 3)] (take the log of both sides)

log y = log 3 . log 10 (log (a^b) = b log a)

log y = log 3

y = 3

but y = 10^(log 3)

.'. 10^(log 3) = 3

 

[vds700]

sorry how can u do the line in bold? is there a specific rule or what? ty

yes i used the log law

log_a xⁿ = n log_a x