Question

Can someone help me with 11 and 12 plss

Answer 1

Q11: x = 3

Q12: x = -4

Explanation:

Q11
We have the equation as

`\:64^(2x\:+\:4)=16^(5x)`

Convert the bases in terms of the common base 4

64 = 16 x 4 = 4 x 4 x 4 = 4³

16 = 4 x 4 = 4²

Therefore the original equation becomes:

`\mbox {\large 4^(3\left(2x+4\right))=4^(2\cdot \:5x)}`

Since the bases are the same in this equation, the exponents must also be the same

Equation the two exponents we get

3(2x + 4) = 2·5x

==> 6x + 12 = 10x

Move 12 to the right side

6x = -12 + 10x

Move 10x to the left:

6x - 10x = -12

-4x = -12

x = -12/-4 = 3

Answer to Q11

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Q12

We can solve this in a similar manner to Q12

27 = 3 x 3 x = 3³

9 = 3 x 3 = 3²

Plugging these in we get

`\mbox {\large 27^x = 3^(3(x)) = 3^(3x)}`

`\mbox {\large 9^ {(x - 2)} = 3^(2(x-2)) = 3^(2x -4) }\\\\`

Thus we get


`\mbox {\large 3^(3x) = 3^(2x -4) }\\\\`

Equating the exponents we get

`3x = 2x - 4\\\\3x - 2x = -4\\\\x = -4`

ANSWER to 12