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2 years ago

Exponential Form

Mathematics

2 answers:

Snezhnost [94]2 years ago

7 0

6. 1/3
8. -4
9. -3
10. 1/6

klio [65]2 years ago

5 0

Loga n=z ⇔a^z=n

Therefore:
3)log 1/4 1=0 ⇔(1/4)⁰=1
4)log 1/2 32 -5 ⇔(1/2)⁻⁵=2⁵=32

6)
log 27 3=n ⇔ 27^n=3
27=3³
(3³)^n=3 ⇒n=1/3 (3³)¹/³=3³/³=3¹=3

Answer: log 27 3=1/3

7)
log1/4 256=n ⇔(1/4)^n=256
256=2⁸=4⁴
(1/4)^n=4⁴ ⇒n=-4 [ (1/4)⁻⁴=4⁴=256 ]

Answer: log1/4 256=-4

9)
log 10 0.001=n ⇔ 10^n=0.001
0.001=10⁻³
10^n=10⁻³ ⇒n=-3

answer: log 10 0.001=-3

10)
log 64 2=n ⇔64^n=2
64=2⁶
(2⁶)^n=2 ⇒2(⁶n)=2¹ ⇒ 6n=1 ⇒ n=1/6

answer: log 64 2= 1/6

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anzhelika [568]

<h3>Answer: k = 6+10x</h3>

=======================================================

Work Shown:

$64*4^{5x}\\\\\left(2^6\right)*\left(2^{2}\right)^{5x}\\\\\left(2^6\right)*\left(2^{2*5x}\right)\\\\\left(2^6\right)*\left(2^{10x}\right)\\\\2^{6+10x}\\\\$

The result is in the form $2^k$ with k = 6+10x

------------------------

Explanation:

First we need to get everything as an exponential expression with a base 2.

We rewrite 64 as $2^6$ and 4 as $2^2$ in the second step.

In the third step, I used the rule $(a^b)^c = a^{b*c}$ which says to multiply the exponents together. That's how I went from $(2^2)^{5x}$ to $2^{2*5x}$ . Then the 2*5x in the exponent becomes 10x.

To wrap things up, I used the rule $a^b*a^c = a^{b+c}$ which says to add the exponents when we multiply stuff of the same base together.

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