Question

Please no dumb answers. I really do need help. And correct answers.
f(x)=-32(2)^(x-3)+3 find the y-intercept?

What is the exponential form of the logarithmic equation?
4=log0.80.4096

What is the value of the logarithm?
log(750)

Round the answer to the nearest thousandth.
Solve the logarithmic equation.
y=log40.25

Answer 1

y intercept is where x=0

so
-32(2^-3)+3
-32(1/(2^3))+3
-32(1/8)+3
-4+3
-1
yint is (0,-1)



log thing

assuming you meant $4=log_{0.80}(0.4096)$ (because I can't read what that is supposed to be )
remember that $a=log_b(c)$ tarnalsates to $b^a=c$ so den
$4=log_{0.80}(0.4096)$ means $4^{0.4096}=0.80$





log(750)
when no base is mentinoed, assume base of 10
I would just use my calculator tho
$log_{10}(750)=?$
translate
$10^?=750$
I do know that 2<?<3 tho, that's all I know
if we did use a calculator then the aproxamate value is 2.875






y=log40.25
10^y=40.25
know taht 1<y<2
dunno what else
sorry
if we use calcualtor it's about 1.604

Answer 2

Answer:

1. y-intercept is -1.

2. Exponential form is $0.80^{4}=0.4096$.

3. 2.875

4. y = 0.873

Step-by-step explanation:

1. We have the function $f(x) = -32 \times 2^{x-3} +3$.

Now as we know that y-intercept is the point where the line cuts y-axis i.e. x=0.

So, x=0 in the $f(x) = -32 \times 2^{x-3} +3$ gives,

$f(x) = -32 \times 2^{0-3} +3$

i.e. $f(x) = \frac{-32}{2^{3}}+3$

i.e. $f(x) = \frac{-32}{8}+3$

i.e. $f(x) = -4+3$

i.e. $f(x) = -1$

So, the y-intercept of this function is -1.

2. We have, $4=\log_{0.80}{0.4096}$

As, $a=\log_{b}c$ implies $b^{a}=c$.

Therefore, $4=\log_{0.80}0.4096$ implies $0.80^{4}=0.4096$.

Hence, the exponential form is $0.80^{4}=0.4096$.

3. We have to find the value of log(750).

As no base is given, we assume the base to be 10.

So, $\log_{10}750$ = 2.875.

4. We have, $y=\log_{40}{25}$

Again using, As, $a=\log_{b}c$ implies $b^{a}=c$.

We have, As, $y=\log_{40}25$ implies $40^{y}=25$

i.e. $y \times \log40 = \log25$

i.e. y =  0.87258905175

Rounding to nearest thousand gives y = 0.873.