solve each equation (algebra 2)?

Question

log4 (little 4 down) x = 3
a. 81
b. 64

log 9 (little 9 down) 27 = x
a. 3/2
b. - 3/2

Answer 1

Question 1
log4 x = 3
x = 4³
x = 64
OPTION b

Question 2
log9 27 = x
9^x = 27
x log 9 = log 27
x = log 27 / log 9
x = log 3³ / log 3²
x = ( 3 log 3 ) / ( 2 log 3 )
x = 3 / 2

Answer 2

Let's have a look at the definition of a logarithm:

I will mark this little number down as {b}.

log{b}a = c such c that b^c = a
example:
log{2}8 = 3 because 2^3 = 8

Take it from here, I'm sure you can do it and good luck,

Kempos

Answer 3

1) b since 4^3 = 64
2) a since 9^(3/2) = 27

Answer 4

Ah, ye olde logarithims...I hate 'em too...

1)put into exponential form: x=4^3 (^3, as in third power), so
x=64...so the answer's b!

2)use change-of-base formula...so, you divide log 27 by log 9 (not little 9, I mean it as in log base 10 of 9)...gives you x=1.5, so the answer's a!

Answer 5

a million. a³-9a²+14a=0 a(a-7)(a-2)=0 solutions are a=0,2 and 7. 2. f(x)=x^4 -14x²+forty 5 if x=3,then x^4 -14x²+forty 5=0,so (x-3) is area of f(x) (x-3)(ax^3+bx^2+cx+d)=x^4 -14x²+forty 5 You get a=a million,b=3,c= -5 and d= -15 yet I depart it to you in looking the different cost/s of x.I provide you one hint: set g(x)=x^3+3x^2-5x-15 and detect a cost of x that makes g(x)=0 and stick to the comparable approach I confirmed you in the previous 3. f(x)=x^4-13x²+36 if x=2,then f(x)=0,so (x-2) is area of f(x) (x-3)(ax^3+bx^2+cx+d)=x^4-13x²+36 You get a=a million,b=2,c= -9 and d= -18 Now g(x)=x^3+2x^2-9x-18 if x= -2,g(x)=0,so (x+2) is area of g(x) (x+2)(kx^2+lx+m)=x^3+2x^2-9x-18 You get ok=a million,l=0 and m= -9,so the different factors of g(x) are (x+3) and (x-3).now you could infer what the values of x could be. 4. x³-3x²=0 (x^2)(x-3)=0 now you could infer what the values of x could be.

Answer 6

4^3 = x
64 = x

9^x = 27
x = 3/2

Answer 7

yea what he said