Please help solve for V1/V2 for 3=20log_10(V2/V1)?
回答 (3)
2017-08-09 12:06 pm
3 = 20 log₁₀(V₂/V₁)
Rearrange, we get :
20 log₁₀(V₂/V₁) = 3
log₁₀(V₂/V₁) = 3/20
By the definition of logarithm: If log₁₀(a) = b, then a = 10^b
Now, log₁₀(V₂/V₁) = 3/20
Then, (V₂/V₁) = 10^(3/20)
If a = b, then 1 / a = 1 / b :
Then, 1 / (V₂/V₁) = 1 / 10^(3/20)
1 × (V₁/ V₂) = 1 / 10^(3/20)
Hence, (V₁/ V₂) = 1 / 10^(3/20)
Rearrange, we get :
20 log₁₀(V₂/V₁) = 3
log₁₀(V₂/V₁) = 3/20
By the definition of logarithm: If log₁₀(a) = b, then a = 10^b
Now, log₁₀(V₂/V₁) = 3/20
Then, (V₂/V₁) = 10^(3/20)
If a = b, then 1 / a = 1 / b :
Then, 1 / (V₂/V₁) = 1 / 10^(3/20)
1 × (V₁/ V₂) = 1 / 10^(3/20)
Hence, (V₁/ V₂) = 1 / 10^(3/20)
2017-08-09 1:22 pm
Let L denote log(base 10). 20L(v2/v1)=3, ie., L(v2/v1)=3/20, ie.10^[L(v2/v1)]=(v2/v1)=10^(0.15)=
1.161834243.
1.161834243.
2017-08-09 11:58 am
3 = 20log(V2/V1)
3/20 = log(V2/V1)
10^(3/20) = V2/V1
3/20 = log(V2/V1)
10^(3/20) = V2/V1
收錄日期: 2021-04-18 17:36:37
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