Question

Please help me with this problem!!!

Answer 1

$3 {}^{4x - 5} = ( \frac{1}{27} ) {}^{2x + 10} \\ note \: \: that = \\ ( \frac{1}{27} ) = ( \frac{1}{3 {}^{3} } ) = 3 {}^{ - 3} \\ \\ 3 {}^{4x - 5} = (3 {}^{ - 3} ) {}^{2x + 10} \\ 3 {}^{4x + 5} = 3 {}^{ - 3(2x + 10)} \\ equal \: \: bases = > equal \: exponents$

$4x - 5 = - 3(2x + 10) \\ 4x - 5 = - 6x - 30 \\ 4x + 6x = - 30 + 5 \\ 10x = - 25 \\ x = - 2.5$

Answer 2

Answer:

x = -2.5

Step-by-step explanation:

3^(4x-5)=(1/27)^(2x+10)

We are working with exponents

We can rewrite 1/27 as 3^-3

3^(4x-5)=(3^-3)^(2x+10)

Now we know that a^b^c = a^(b*c)

3^(4x-5)=3^( -3* (2x+10))

3^(4x-5)=3^( -6x -30)

The bases are the same so the exponents must be the same

4x-5 = -6x-30

Add 6x to each side

4x-6+6x = -6x-30+6x

10x-6 = -30

Add 5 to each side

10x-5+5 = -30 +5

10x = -25

Divide by 10

10x/10 = -25/10

x = -2.5