Solve the inequality. 64^x<32^(x+2).

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

64^x < 32^(x+2).

First we will re-write the numbers as a power.

We know that"

64 = 2*2*2*2*2*2 = 2^6

32= 2*2*2*2*2 = 2^5

Let us substitute:

(2^6)^x < (2^5)^(x+2)

From exponent properties we know that:

a^b^c = a^(bc)

==> 2^(6x) < 2^(5x+10)

==> 2^6x = 2^5x+10

Now, we have the bases are equal, then the powers should satisfy the inequality.

==> 6x < 5x + 10

We will subtract 5x from both sides.

==> x < 10

Then the answer is:

x  belongs to the interval ( -inf, 10)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that 64 and 32 are powers of 2, so we'll change the bases 64 and 32 in powers of 2.

64 = 2^6

32 = 2^5

We'll re-write the inequality:

(2^6)^x < (2^5)^(x+2)

2^6x < 2^5(x+2)

Since the bases are > 1, then the exponential function is increasing and we'll apply one to one property, keeping the direction of the inequality:

6x < 5(x+2)

We'll remove the brackets:

6x < 5x + 10

6x - 5x < 10

x < 10

The interval of admissible values of x is: (-infinite ; 10).

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