Exponential and Logarithmic Problem

[JoeC]

I am looking for help solving for x for the question below. Any help would be greatly appreciated.

2^x=16(8^2x)

 

[Ackbach]

I am looking for help solving for x for the question below. Any help would be greatly appreciated.

2^x=16(8^2x)

What have you tried? Where are you stuck?

 

[JoeC]

I don't really know where to start with this.

- - - Updated - - -

the answer to the question is -4/5 but I haven't been able to get it using log or ln.

 

[I like Serena]

I am looking for help solving for x for the question below. Any help would be greatly appreciated.

2^x=16(8^2x)

I don't really know where to start with this.

- - - Updated - - -

the answer to the question is -4/5 but I haven't been able to get it using log or ln.

Welcome to MHB, JoeC! :)

You have
$$2^x=16(8^{2x})$$
(Or at least that is what I assume you have.)

Since we have $2^x$, let's try to make the other power also a power of $2$.
We have that $8=2^3$.
Do you know what $(2^3)^{2x}$ is?

 

[CellCoree]

To solve for x in this exponential and logarithmic problem, we can use the properties of exponents and logarithms. First, we can rewrite 16 as 2^4. Then, using the power rule of exponents (a^b)^c = a^(bc), we can rewrite 8^2x as (2^3)^2x = 2^(3*2x) = 2^6x. Substituting these values into the original equation, we get:

2^x = 2^4 * 2^6x

Using the product rule of exponents (a^b * a^c = a^(b+c)), we can combine the bases of 2 on the right side of the equation:

2^x = 2^(4+6x)

Now, using the fact that if a^x = a^y, then x = y, we can set the exponents on both sides of the equation equal to each other:

x = 4+6x

Solving for x, we get x = -1/5. Therefore, the solution to the exponential and logarithmic problem is x = -1/5.