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Inessa [10]
4 years ago
14

Which equation can be solved using the inverse property?

Mathematics
2 answers:
Ganezh [65]4 years ago
8 0

Answer:

The equation which can be solved by inverse property is: log_{2}x=log_{2} 6.

Step-by-step explanation:

The inverse property of Logarithms:

log_{b}b^{x}=x

or

g(x)=b^{x} \ \text{and} \ \ f(x)=log_{b} x

From the provided option log_{2}x=log_{2} 6 can be solved using the inverse property.

Consider the option 1. log_{2}x=log_{2} 6

Use the inverse property of logarithms:

2^{x}=2^{6}

x=6

Thus, log_{2}x=log_{2} 6 can be solved by inverse property.

Kazeer [188]4 years ago
4 0
The inverse property of a logarithm is
loga a^x = x
or
a^loga x = x
Therefore,from the choices, the equation than can be solved using the inverse property is
<span>log2x = log26
</span><span>which if simplified and solved results to
x = 6</span><span />
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In △ABC, point P∈AC with AP:PC=1:3, point Q∈AB so that AQ:QB=3:4, Find the ratios A(PBQ) : A(PBC) and A(AQP) : A(ABC)
ludmilkaskok [199]

Answer:

The ratio APBQ : APBC is 4:21, AAQP : AABC is 3:28

Step-by-step explanation:

Data provided in the question

AP : PC = 1 : 3

So let us assume AP = x and PC = 3x.

And, If AQ : QB = 3 : 4

So, let us assume AQ = 3y and QB = 4y.

Now we have to find the area of ΔAQP and ΔABC

A_{AQP}=\dfrac{1}{2}\cdot AP\cdot AQ\cdot \sin\angle A=\dfrac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A;

A_{ABC}=\dfrac{1}{2}\cdot AC\cdot AB\cdot \sin\angle A=\dfrac{1}{2}\cdot (x+3x)\cdot (3y+4y)\cdot \sin\angle A=\dfrac{1}{2}\cdot 4x\cdot 7y\cdot \sin\angleA

Therefore

\dfrac{A_{APQ}}{A_{ABC}}=\dfrac{\frac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A}{\frac{1}{2}\cdot 4x\cdot 7y\cdot \sin\angleA}=\dfrac{3}{28}.

Now

\dfrac{A_{APQ}}{A_{ABP}}=\dfrac{\frac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A}{\frac{1}{2}\cdot x\cdot (3y+4y)\cdot \sin\angle A}=\dfrac{3}{7}.

Now after solving these two ratios we can find

A_{ABP}=\dfrac{7}{3}A_{APQ}\Rightarrow A_{PBQ}=A_{APB}-A_{APQ}=\dfrac{7}{3}A_{APQ}-A_{APQ}=\dfrac{4}{3}A_{APQ}

A_{ABC}=\dfrac{28}{3}A_{APQ}\Rightarrow A_{PBC}=A_{ABC}-A_{APB}=\dfrac{28}{3}A_{APQ}-\dfrac{7}{3}A_{APQ}=7A_{APQ}.

Therefore

\dfrac{A_{PBQ}}{A_{PBC}}=\dfrac{\frac{4}{3}A_{APQ}}{7A_{APQ}}=\dfrac{4}{21}.

Hence, we applied the above equation so that we can get to know the ratios

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