MHB Exponential and Logarithmic Problem

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The discussion focuses on solving the equation 2^x = 16(8^2x) for x. The user is struggling to start and has not successfully used logarithms to find the solution, which is known to be -4/5. Another participant suggests rewriting the equation using powers of 2, noting that 8 can be expressed as 2^3. The conversation emphasizes the need to convert all terms to a common base to facilitate solving the equation.
JoeC
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I am looking for help solving for x for the question below. Any help would be greatly appreciated.

2^x=16(8^2x)
 
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JoeC said:
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?
 
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.
 
JoeC said:
I am looking for help solving for x for the question below. Any help would be greatly appreciated.

2^x=16(8^2x)

JoeC said:
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?
 

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