Solve each of the following without using a calculator: 4^(2x) = 64^(x-2)?

Question

4^(2x) = 64^(x-2)


log2x + log2(x+3) = log2 18

log2 (1/8)

(please note that those log2 is really a log with a subscript 2.


Thanks, I have alot more problems, but these are the basics, once I see how to do these I will probably be good.

Answer 1

4^2x = 64^(x-2)

what you want to do is create bases that are equal, in this case, bases of 2 will be fine (the smallest base possible makes it easier to calculate)

2^(2*2x) = 2^6(x-2)

as the bases are equal, the powers are also equal. there fore,
4x=6x-12
or x = 6


log2x + log2(x+3) = log2 18
or, log2(x*(x+3)) = log2 18

now, equating the values,

x*(x+3) = 18
or, x^2 + 3x = 18
or, x^2 + 3x -18 = 0
or, x^2 + 6x - 3x - 18 = 0
or, x(x+6) -3(x+6) = 0
or (x-3)(x+6)=0

therefore, x = 3, -6
However, if x=-6, then the value of log2 x is not defined (values of logs of -ve numbers do not exist), hence x = 3

log2(1/8)= log2((1/2)^3)
= log2 (2^-3)
= -3

Answer 2

4^(2x) =64^(x-2) ; 64 = 4^3 so 64^(x-2) = 4^(3x-6)
2 numbers with same base are equal if exponents are equal
2x = 3x-6
Answer x=6 check it works

log2 x +log2 (x+3 ) = log2 18 log a +log b = log ab so
x(x+3) =18---> x^2+3x-18 =0 ANswer x=3

log2 (1/8) = log2 (1/2^3)= log2 (2^(-3))

answer -3

Answer 3

If you ever took precalculus or some other math class which covers properties of functions, you learn how to determine the domain of the function. One of the red flags to look out for is when you divide by zero. In your expression, when x = 3, the denominator is zero. Thus, the domain is every number beside 3. Unfortunately, the number you are trying to plug in (x = 3) is the only number that doesn't work in this function. To see this first hand, you0 can graph this function on a TI-83. If you zoom in on the point of the graph at x = 3, you will see that there is a blank spot there! That is because, as stated above, there just isn't a value of the expression at x = 3. You may say, well it looks like the answer should be 6, looking at the graph. This concept of what the answer "should be" is what limits are all about. The values of the function on the left and right of x = 3 all go towards 6 as you get closer and closer. So we say the limit as x goes to 3 is 6. So even though it is not technically the answer, 6 is your best choice. Zero is absolutely not correct in any sense. The very best answer is to say that the expression is undefined at x = 3. This problem illustrates why 0/0 is called indeterminate. In this problem, 0/0 in a way equals 6. The idea that 0/0 can equal anything is actually the essence of calculus.

Answer 4

1). 4^(2x) = 64^(x-2)
Notice 64 = 4^3,
So, 4^(2x) = 4^[3*(x-2)]
Now, since the base are the same, we can just ignore them,
And now we have: 2x = 3*(x-2)
2x = 3x - 6
0 = x - 6
Therefore, x = 6

2). log2x + log2(x+3) = log2 18
Using log rules ==> loga (x) + loga (y) = log (x*y); [NB: Sub-scipt 'a']
So, log2x + log2(x+3) = log2 [x*(x+3)] = log2 (x^2+3x)
Now we have: log2 (x^2+3x) = log2 (18)
We can ignore the log2 again since they are the same,
Hence, x^2+3x = 18
x^2+3x-18=0 (Quadratic!)
(x+3)(x-6)=0
Therefore, x=-3 or x=6

3). log2 (1/8)
This time use log rule ==> loga (x/y) = loga (x) - loga (y) [Subscript 'a' again.]
So it's log2 (1) - log2 (8) [NB: loga (1)=0, and 8=2*4=2*(2^2)]
= 0 - [log2 (2*2^2) [Remember the previous log rule in 2).?]
= 0 - [log2 (2) + 2log2 (2)] [NB: loga (x^z)=z*loga (x)]
= 0 - (1 + 2*1) [For loga (a) = 1]
= 0 - 1 - 2
= -3

Answer 5

Since 64=4^3, the first can be rewritten and 4^(2x)=4^[3(x-2)]
so 2x=3x-6--or x=6.

The second one can be rewritten as log2[x(x+3)]=log218
so x(x+3)=18 or x^2+3x-18=0 so (x+6)(x-3)=0 so x=3 only.

Answer 6

4^(2x)=64^(x-2)
=>4^(2x)=4^(3x-6)
=>2x=3x-6
=>x=6 ans

log2(x)+log2(x+3)=log2 18
=>log2 x*(x+3)=log2 18
=>x(x+3)=18
=>x^2+3x-18=0
=>x^2+6x-3x-18=0
=>x(x+6)-3(x+6)=0
=>(x+6)(x-3)=0
=>x=-6 or 3

log2 (1/8)=log2(1)-log2(8)
=0-log2(2^3)
= - 3 log2 2
=-3*1
=-3

Answer 7

4^2x = 64^(x-2) = 4^3(x-2) as 64 = 4^3

so we get 2x = 3x - 6 or x = 6

log (base 2) 1/8 = -3 as 2^3 =8 or 2^(-3) = 1/8

log 2 (x) + log 2 (x+3) = log 2 18

so x (x+3) = 18 as log a + log b = ab
u can solve it as x = 3 or -6

-6 is not possible so ans 3

Answer 8

4^(2x) = 64^(x-2)
4^(2x) = (4^3)^(x-2)
4^(2x) = 4^(3x-6) .............2x=3x-6 ............x=6

log2x + log2(x+3) = log2 18 =log2 x (x+3) = log2 18
x(x+3) =18 ..............x^2 +3x -18 = 0 ...........(x+6)(x-3) =0
x= -6 wrong x =3

log2 (1/8)= log2 (2^ -3) = -3 log2 2 = -3

Answer 9

4^2x=64^(x-2)=(4^3x)/(4^6)
=>(4^3x)/(4^2x)=(4^6)
=>(4^x)=(4^6)
since base is equal x=6;
log(x/x+3)=log 18 =>x/x+3=18=>x=18x+51=>17x=-51
=>x=-3;
log2(1/8)=log2(2^-3)
=>-3log2(2)=-3(1)=-3;