Solve 8^x+4^x=32

Solve 8^x+4^x=32

Postby Guest » Wed Aug 04, 2021 6:12 pm

Solve [tex]8^x+4^x=32[/tex]
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Re: Solve 8^x+4^x=32

Postby Guest » Thu Aug 05, 2021 12:01 pm

[tex]8= 2^3[/tex] so [tex]8^x= (2^3)^x= 2^{3x}= (2^x)^3[/tex].
[tex]4= 2^2[/tex] so [tex]4^x= (2^2)^x= 2^{2x}= (2^x)^2[/tex].

So the equation is [tex](2^x)^3+ (2^x)^2= 32[/tex]

Let [tex]y= 2^x[/tex]'
Then the equation becomes [tex]y^3+ y^2= 32[/tex].

I don't see any elementary way to solve that. Graphing shows that there is a solution slightly less than y= 3 so x is slightly less than log 3/log 2.
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Re: Solve 8^x+4^x=32

Postby Guest » Fri Aug 06, 2021 3:48 pm

After reducing to [tex]y^3+ y^2= 32[/tex] perhaps a numerical method. If y= 3 [tex]x^3+ x^2= 27+ 9= 36[/tex] which is a little larger than 32, If y= 2,8 [tex]x^3+ x^2= 21.952+ 7.84= 29.692[/tex] which is a little less than 32. So try half way between, y= 2.9. If y= 2.9 [tex]x^3+ x^2= 32.799[/tex] which is a little larger than 32 so a root lies between 2.8 and 2.9, 2.85 say. [tex]x^3+ x^2= 31.271625[/tex]. That is a little less than 32 (but much close than before) so a root lie between 2.85 and 2.9. Taking half way between again, try y= 2.875. Continue in that way until you get the accuracy you wish.
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Re: Solve 8^x+4^x=32

Postby Guest » Fri Mar 29, 2024 8:10 am

To solve the equation 8^x+4^x=32, we can recognize that both 8 and 4 are powers of 2.

Rewriting them in terms of 2, we have 2^3x+2^2x=2^5.

Now, notice that 32=2^5. So, we can rewrite the equation as 2^3x+2^2x=2^5.

This gives us 2^2x(2^x+1)=25.

Dividing both sides by 2^2x, we get 2^x+1=2^(5−2x).

Now, we see that the left-hand side is increasing while the right-hand side is decreasing as x increases.

Therefore, there's only one solution to this equation. By inspection, it's evident that x=1 satisfies the equation.

Hence, 8^1+4^1=32, fulfilling the condition.
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Re: Solve 8^x+4^x=32

Postby Guest » Fri Mar 29, 2024 8:12 am

To solve the equation 8^x+4^x=32, we can recognize that both 8 and 4 are powers of 2. Rewriting them in terms of 2, we have 2^3x+2^2x=2^5.
Now, notice that 32=2^5. So, we can rewrite the equation as 2^3x+2^2x=2^5.
This gives us 2^2x(2^x+1)=25.
Dividing both sides by 2^2x, we get 2^x+1=2^(5−2x).
Now, we see that the left-hand side is increasing while the right-hand side is decreasing as x increases.
Therefore, there's only one solution to this equation. By inspection, it's evident that x=1 satisfies the equation.
Hence, 8^1+4^1=32, fulfilling the condition


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