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Sagot :
The goal is to make the bases the same and so 27 can become 3^3
3^5x = 3 ^ 3(2x-4)
This becomes
3^5x= 3 ^6x-12
Now that the bases are the same, set the exponents equal to one another
5x= 6x-12
Subtract 6x on both sides
-x= -12
Divide -1 on both sides
X=12
#brainliest????
3^5x = 3 ^ 3(2x-4)
This becomes
3^5x= 3 ^6x-12
Now that the bases are the same, set the exponents equal to one another
5x= 6x-12
Subtract 6x on both sides
-x= -12
Divide -1 on both sides
X=12
#brainliest????
Answer:
x = 12
Step-by-step explanation:
First:
[tex]27=9*3=3*3*3=3^3[/tex]
If you rewrite the 27 into base 3 like that:
[tex]3^{5x}=3^{(3)\times(2x-4)[/tex]
Now that both sides have the same base, you can cancel them out. This is the simplest example of that I can think of:
[tex]2^x=2^x[/tex]
Here, the exponent obviously has to be equal for this to be true, right? The exponents do have to be equal, so you can drop the bases completely:
[tex]x=x[/tex]
Same for the problem we have above. It's a base 3 on both sides, and for them to be equal, the exponents also have to be equal.
[tex]3^{5x}=3^{(3)\times(2x-4)}\\5x=(3)\times(2x-4)\\5x=3(2x-4)[/tex]
Now, you can simply solve for x.
[tex]5x=6x-12\\-1x=-12\\x=12[/tex]
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