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postnew [5]
4 years ago
10

How to change the expression to a single logarithm

Mathematics
1 answer:
IrinaK [193]4 years ago
6 0
First, you'll use the "Log power rule," which says that \log_ax^p=p\log_ax. In this case, you're going from the form on the right (with p in front of the log) to the form on the left (with p in the exponent position). So, the expression becomes:

\log_7x^4+\log_7y^8+\log_7z^4

Then, you'll use the "Log product rule," which says that \log_a(xy)=\log_ax+\log_ay. Again, you're going from the form on the right to the form on the left (basically, from the sum of the logs, to a log of the products). So you get:

\log_7(x^4y^8z^4)

There's your expression simplified into a simple logarithm. 
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Rationalise the denominator of:<br>1/(√3 + √5 - √2)​
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Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

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So, using this, we get

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\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3}  +  \sqrt{5}  -  \sqrt{2} } =\dfrac{-  \sqrt{3 \times 3 \times 2} +  \sqrt{2 \times 2 \times 3}  + \sqrt{30}}{12}}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h3><u>More Identities to </u><u>know:</u></h3>

\purple{\boxed{\tt{  {(x  -  y)}^{2} =  {x}^{2} - 2xy +  {y}^{2}}}}

\purple{\boxed{\tt{  {(x   +   y)}^{2} =  {x}^{2} + 2xy +  {y}^{2}}}}

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\purple{\boxed{\tt{  {(x - y)}^{3} =  {x}^{3} - 3xy(x  -  y) -  {y}^{3}}}}

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\pink{\boxed{\tt{  {(x + y)}^{2}  -  {(x - y)}^{2} = 4xy}}}

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