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arsen [322]
3 years ago
13

Expand log_1/2(3x^2/2) using the properties and rules for logarithms. Please help.

Mathematics
2 answers:
Alinara [238K]3 years ago
7 0

Answer:

-0.58-6.64log(x)

Step-by-step explanation:

The given expression is

log_{\frac{1}{2}}(\frac{3x^2}{2})

It can be rewritten as

log_{\frac{1}{2}}(\frac{1}{2}\times 3\times x^2)

Using the properties and rules for logarithms, we get

log_{\frac{1}{2}}(\frac{1}{2})+log_{\frac{1}{2}}(3)+log_{\frac{1}{2}}(x^2)         [\because log_a(mn)=log_am+log_an]

1+log_{\frac{1}{2}}(3)+2log_{\frac{1}{2}}x         [\because log_a(a)=1,logx^n=nlogx]

1+\frac{log(3)}{log(\frac{1}{2})}+2\frac{log(x)}{log(\frac{1}{2})}             [\because log_a(b)=\frac{logb}{loga}]

1+\frac{log(3)}{log(1)-log(2)}+2\frac{log(x)}{log(1)-log(2)}        [\because log\frac{a}{b}=loga-logb]

We know that,

log (1) = 0

log (2) = 0.301

log (3) = 0.477

Substitute these values in the

1+\frac{0.477}{0-0.301}+2\frac{log(x)}{0-0.301}

1-1.58-2\frac{log(x)}{0.301}

-0.58-6.64log(x)

Therefore, the expanded form of given expression is -0.58-6.64log(x).

Naily [24]3 years ago
6 0

Answer:

\frac{0.1761+2\log(x)}{-0.3010}

Step-by-step explanation:

Data provided:

\log_{1/2}(\frac{3x^2}{2})

now,

we know the properties of log functions as:

1) log(AB) = log(A) + log(B)

2) \log(\frac{A}{B}) = log(A) - log(B)

3) log(xⁿ) = n × log(x)

thus,

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3x^2) - \log_{1/2}(2)

or

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3)+\log_{\frac{1}{2}}(x^2) - \log_{1/2}(2)

or

using property 3

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3)+2\log_{\frac{1}{2}}(x) - \log_{1/2}(2)

also,

\log_a(x)=\frac{\log(x)}{\log(a)}

thus,

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)}{\log(\frac{1}{2})}+2\times[\frac{\log(x)}{\log(\frac{1}{2}}]-\frac{\log(2)}{\log(\frac{1}{2})}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)+2\log(x)-\log(2)}{\log(\frac{1}{2})}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)+2\log(x)-\log(2)}{\log(1)-log(2)}

now,

log(1) = 0

log(2) = 0.3010

log(3) = 0.4771

thus,

\log_{1/2}(\frac{3x^2}{2}) = \frac{0.4771+2\log(x)-0.3010}{0-0.3010}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{0.1761+2\log(x)}{-0.3010}

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