Question

Solve 3^(4x-5)=〖27〗^(x+10) for x. Show your work

Answer 1

Answer:

x =35

Step-by-step explanation:

3^(4x-5)=〖27〗^(x+10)

We need to get the same base

27 = 3^3

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

We know a^b^c = a^(b*c)

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

Since the bases are the same, the exponents are equal

4x-5 = 3(x+10)

Distribute

4x-5 = 3x+30

Subtract 3x from each side

4x-3x-5 = 3x-3x+30

x-5 =30

Add 5 to each side

x-5+5 = 30+5

x = 35

Answer 2

Answer:

The answer is x=35

Step-by-step explanation:

Right Side:

First we convert $27^{x+10}\mathrm{\:to\:base\:}3$. This result in $\left(3^3\right)^{x+10}$.

Left Side:

We can convert the left side the same way: $3^{4x-5}=3^{3\left(x+10\right)}$

Both Sides:

Since both sides have the same base, we can set their exponents equal to each other.

4x-5 = 3(x+10).

At this point we can solve for x normally and get x = 35.