How can you solve $5^{x^2+8}=125^{2x}$ for x?

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  • Thread starter karush
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In summary, To solve $5^{x^2+8}=125^{2x}$, we first use the fact that they have a common base of 5. Then, we can rewrite $125^{2x}$ as $(5^3)^{2x}$, which equals $5^{6x}$. Next, we equate the exponents and get $x^2+8=6x$. This can be factored to $(x-2)(x-4)=0$, giving us the solutions $x=2$ and $x=4$. Finally, checking both values shows that they are indeed correct.
  • #1
karush
Gold Member
MHB
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5
$\tiny [6.3.74 Miliani HS$
Find x
$5^{x^2+8}=125^{2x}$
$\begin{array}{rlll}
\textsf{common base}&125^{2x}=(5^3)^{2x}=5^{6x}\\
\textsf{then } &x^2+8=6x\implies x^2-6x+8=0 \\
\textsf{factor}&(x-2)(x-4)=0\\
\textsf{get zeros}&x=2, \quad x=4\\
\end{array}$

should be ok
suggestions...
 
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  • #2
that’s fine
 
  • #3
I would suggest that you CHECK your answers!

If x= 2 then $x^2+ 8= 4+ 8= 12$ so $5^{x^2+ 8}= 5^{12}= 244140625$ while $125^{2x}= 125^{4}= 244140625$. Yes, they are equal!

If x= 4
then $x^2+ 8= 16+ 8= 24$ so $5^{x^2+ 8}= 5^{24}= 59604644775390625$ while $125^{2x}= 125^{8}= 59604644775390625$. Yes, they are equal!
 
  • #4
If you live in Hawaii, how can you bear to do mathematics rather than being out on the beach every day?
 
  • #5
there aways seems to be some non textbook trick with logs
 
  • #6
Beer soaked ramblings follow.
Country Boy said:
If you live in Hawaii, how can you bear to do mathematics rather than being out on the beach every day?
Not everyone is fond of the sun and the beach.
Some would rather be indoors away from pesky flies and mosquitoes.
 
  • #7
jonah said:
Beer soaked ramblings follow.

Not everyone is fond of the sun and the beach.
Some would rather be indoors away from pesky flies and mosquitoes.
im 76 and over weight I am embarrased to be seen in swimsuit
besides the mask restrictions have been ridiculus here. but they are getting ignored more and more finally,,
now they wondering if we will have condo colapse like florida :confused:
 
  • #8
Country Boy said:
I would suggest that you CHECK your answers!

If x= 2 then $x^2+ 8= 4+ 8= 12$ so $5^{x^2+ 8}= 5^{12}= 244140625$ while $125^{2x}= 125^{4}= 244140625$. Yes, they are equal!

If x= 4
then $x^2+ 8= 16+ 8= 24$ so $5^{x^2+ 8}= 5^{24}= 59604644775390625$ while $125^{2x}= 125^{8}= 59604644775390625$. Yes, they are equal!
actually I am more interested in the steps
probably don't need the decimal unless there is some purpose for it
I ussually check with W|A if the book does not give answers
 
  • #9
karush said:
actually I am more interested in the steps
probably don't need the decimal unless there is some purpose for it
I ussually check with W|A if the book does not give answers
Never trust the calculator!

-Dan
 

FAQ: How can you solve $5^{x^2+8}=125^{2x}$ for x?

What does the equation "6.3.74 5^{x^2+8}=125^{2x}" represent?

The equation represents an exponential function where the base on the left side is 5 and the base on the right side is 125. The exponents are both expressions involving the variable x.

How do I solve for x in this equation?

To solve for x, you can take the logarithm of both sides with the same base. This will allow you to bring down the exponents and solve for x algebraically.

Can I use any base for the logarithm?

Yes, as long as the base is positive and not equal to 1. However, using the same base as the exponential function will result in a simpler solution.

Is there a specific method for solving exponential equations?

Yes, there are various methods such as taking logarithms, using the change of base formula, or using the properties of exponents. It is important to choose a method that is most suitable for the given equation.

Can this equation be solved without using logarithms?

Yes, it is possible to solve this equation without using logarithms by using the properties of exponents and simplifying both sides until the variable can be isolated. However, using logarithms may result in a quicker and more efficient solution.

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