Solve the exponential equation by rewriting the base. explain the steps towards your answer.

The answer is x = -0.4

Step-by-step explanation:

* In the exponential functions we have some rules

1- b^m  ×  b^n  =  b^(m + n) ⇒ in multiplication if they have same base we add  the power

2- b^m  ÷  b^n =  b^(m – n) ⇒  in division if they have same base we subtract  the power

3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply them

4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same  power then we multiply them ad put over the answer the power

5- b^(-m) = 1/(b^m)  (for all nonzero real numbers b) ⇒ If we have negative power we reciprocal the base to get positive power

6- If  a^m  =  a^n  ,  then  m  =  n ⇒ equal bases get equal powers

7- If  a^m  =  b^m  ,  then  a  =  b    or    m  =  0

* Now lets solve our problem

∵ 32^(-2x) = 16

∵ 32 = 2^5 ⇒ 32 ÷ 2 =16 ÷ 2 = 8 ÷ 2 = 4 ÷ 2 = 2 ÷ 2 = 1

∵ 16 = 2^4 ⇒ 16 ÷ 2 = 8 ÷ 2 = 4 ÷ 2 = 2 ÷ 2 = 1

∵ (2^5)^(-2x) = 2^(5 × -2x) = 2^(-10x) ⇒ by using rule 3

∴ 2^(-10x) = 2^4 ⇒ by using rule 6

∴ -10x = 4 ⇒ divided by -10 for both sides

∴ x = 4/-10 = -2/5 = -0.4

* The answer is x = -0.4

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.