Which equation is equivalent to????

Mathematics

1 answer:

vesna_86 [32]3 years ago

3 0

3² = x + 5. The equivalent equation to log₃(x+5) = 2 is 3² = x + 5.

In order to solve this problem we have to know about Logarithm.

Logarithm is defined as the exponent of a power with a certain base, that is, the number to which a given base must be raised to obtain a given result.

logₐ b = x ----------> aˣ = b

Solving for log₃(x+5) = 2 where a = 3, x = 2, and b = x + 5

3² = x + 5

You might be interested in

Find the unit vector in the direction of u = (-3,2).

DENIUS [597]

$\bf \textit{unit vector for }(a,b)\implies \left( \cfrac{a}{\sqrt{a^2+b^2}}~~,~~\cfrac{b}{\sqrt{a^2+b^2}} \right)\\\\
-------------------------------\\\\
(-3,2)\qquad \stackrel{unit~vector}{\implies }\qquad \left( \cfrac{-3}{\sqrt{(-3)^2+2^2}}~~,~~\cfrac{2}{\sqrt{(-3)^2+2^2}} \right)
\\\\\\
\left( -\cfrac{3}{\sqrt{13}}~~,~~ \cfrac{2}{\sqrt{13}}\right)\\\\
-------------------------------$

$\bf \textit{and now let's \underline{rationalize} the denominator for each}
\\\\\\
-\cfrac{3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies -\cfrac{3\sqrt{13}}{13} \qquad \qquad \qquad \qquad \cfrac{2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{2\sqrt{13}}{13}
\\\\\\
\textit{and written in \underline{ai+bj form}}\qquad -\cfrac{3\sqrt{13}}{13}i~~~~+~~~~\cfrac{2\sqrt{13}}{13}j$

7 0

3 years ago

Quadrilateral ABCD has vertices A(-3, 4), B(1, 3), C(3, 6), and D(1, 6). Match each set of vertices of quadrilateral EFGH with t

Maru [420]

Answer:

See explanation

Step-by-step explanation:

First plot points A, B, C and D on the coordinate plane and connect them to get quadrilateral ABCD. Then plot the vertices E, F, G, H of the quadrilateral EFGH in each case.

For each of these cases there are attached diagrams

1 case - reflection acraoss the x-axis

2 case - translation 5 units down

3 case - reflection across the y-axis

4 case - translation 7 units right