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Lina20 [59]
3 years ago
9

Expand the following logs:

1" title="log_{2} \sqrt{ab^{3} }" alt="log_{2} \sqrt{ab^{3} }" align="absmiddle" class="latex-formula">

SHOW ALL WORK.
Mathematics
1 answer:
aev [14]3 years ago
3 0

Answer:

\log_2(ab^3)^{\frac{1}{2}}=\frac{1}{2}[\log_2(a)+3\log_2(b)].

Step-by-step explanation:

The given logarithmic expression is \log_2\sqrt{ab^3}.

We rewrite the radical as an exponent to obtain;

\log_2(ab^3)^{\frac{1}{2}}.

Recall that; \log_a(M^n)=n\log_a(M)

We apply this rule to obtain;

=\frac{1}{2}\log_2(ab^3).

We now use the rule: \log_a(MN)=\log_a(M)+\log_a(N)

This implies that;

=\frac{1}{2}[\log_2(a)+\log_2(b^3)].

We again apply: \log_a(M^n)=n\log_a(M)

=\frac{1}{2}[\log_2(a)+3\log_2(b)].

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2 years ago
In the figure, triangle CAE is an enlargement of triangle CBD with scale factor of 4/3. The area of the smaller triangle is 9cm2
balandron [24]

Answer:

16 cm^2

Step-by-step explanation:

Given

\triangle CAE -- Bigger Triangle

\triangle CBD -- Smaller Triangle

k = \frac{4}{3} --- Scale factor

Area of CBD = 9

Required

Determine the area of CAE

The area of triangle CBD is:

A_1 = \frac{1}{2}bh

\frac{1}{2}bh = 9

The area of CAE is:

A_2 = \frac{1}{2}BH

Where:

B = \frac{4}{3}b and

H = \frac{4}{3}h

The above values is the dimension of the larger triangle (after dilation).

So, we have:

A_2 = \frac{1}{2}*\frac{4}{3}b * \frac{4}{3} * h

A_2 = \frac{1}{2}*\frac{4}{3} * \frac{4}{3} *b* h

A_2 = \frac{1}{2}*\frac{16}{9}  *b* h

Re-order

A_2 = \frac{16}{9}*\frac{1}{2}* b* h

A_2 = \frac{16}{9}*\frac{1}{2}bh

Recall that:

\frac{1}{2}bh = 9

A_2 = \frac{16}{9}*9

A_2 = 16

Hence, the area is 16 cm^2

3 0
3 years ago
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The statement that are  important when representing functions are:

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<h3>What is are functions?</h3>

A function can be described as the  relation between a set of inputs that have  one output each.  it help to display the relationship between inputs in a case whereby  each input is related to exactly one output.

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