Solving Exponential Eqn: 4(2)^5x = sqrt(2)

AND JBLIn summary, to solve the exponential equation (4)(2)^5x=square root of 2, the common base of 2 was found and both sides were multiplied by ^2 to get rid of the square root on the right. However, the exponents were incorrectly added instead of multiplied, resulting in the wrong answer. The mistake was corrected and the final solution was found to be x=-3/10. It is always important to check the solution by substituting it back into the original equation.
  • #1
aisha
584
0
To solve this exponential equation (4)(2)^5x=square root of 2
I know the common base is 2 but how do i make the square root of 2 also a common base? Do i multiply both sides by ^2? When I did this I got x=1/40 I am not sure if this is the right way to do this question can someone tell me :confused:
 
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  • #2
[tex]4*(\sqrt{2} * \sqrt{2})^{5x} = \sqrt{2} [/tex]
 
  • #3
I think you might be on the right track... could you explain what you did? I might have done it the way you were thinking but I got a different answer.
 
  • #4
What I did

ms. confused said:
I think you might be on the right track... could you explain what you did? I might have done it the way you were thinking but I got a different answer.

well first I found the common base of left side and got
(2^2) (2)^5x=square root of 2
then to get rid of the square root on the right I wasnt sure what to do but I decided to multiply both sides by ^2 and got [(2^2) (2)^5x]^2=2 so the exponents were [(2)(5x)]2 = 2 and then when i further simplified I got the exponent x to =1/40
 
  • #5
When you got to here,
aisha said:
[(2)(5x)]2 = 2
, what did you do with the ^2 after the 2?
 
  • #6
ms. confused said:
When you got to here,, what did you do with the ^2 after the 2?

I am not sure what u are talking about. :uhh:
 
  • #7
Write both sides as powers of sqrt(2) and then compare powers.
 
  • #8
lol sorry still confused can someone show me :uhh:
 
  • #9
From here "[(2^2) (2)^5x]^2=2"

to here

"so the exponents were [(2)(5x)]2 = 2"

You ran into some difficulty. The first equation is [tex]\left[2^{2}2^{5x}\right]^2=2[/tex], which is fine. When you combined the terms inside the [] you multiplied the exponents when you should have added them. You should get [tex]\left[2^{2+5x}\right]^2=2[/tex]. Can you manage from here?
 
  • #10
shmoe said:
From here "[(2^2) (2)^5x]^2=2"

to here

"so the exponents were [(2)(5x)]2 = 2"

You ran into some difficulty. The first equation is [tex]\left[2^{2}2^{5x}\right]^2=2[/tex], which is fine. When you combined the terms inside the [] you multiplied the exponents when you should have added them. You should get [tex]\left[2^{2+5x}\right]^2=2[/tex]. Can you manage from here?

Ok I see my mistake but not sure if my answer is right now i got x=-1/5
 
  • #11
No, but I think I see where you're going wrong.
Does this help (I've made a small change) ? [tex]\left[2^{2+5x}\right]^2=2^1[/tex].
 
  • #12
Hold on-remember the power of 2 outside the square brackets [].

It's always a good idea after solving one of these to test your solution back in the original equation.
 
  • #13
:confused: ok this is what I am doing now the exponents are (2+5x)^2=1 this to 4+10x=1 then i get -3/10 =x

oh to check myself where do i sub in x? which part?
 
  • #14
Your original equation was [tex]4*(2)^{5x} = \sqrt{2} [/tex]. If you replace the "x" in this equation with "-3/10", the equation should be true.
 
  • #15
shmoe said:
Your original equation was [tex]4*(2)^{5x} = \sqrt{2} [/tex]. If you replace the "x" in this equation with "-3/10", the equation should be true.

:smile: :rofl: YAYAYA
IM SOO HAPPY LOL I finally got that THANKS SO MUCH everyone ESP SHMOE
 

1. How do I solve an exponential equation like 4(2)^5x = sqrt(2)?

To solve this equation, you first need to isolate the exponential term on one side of the equation. In this case, we can divide both sides by 4 to get (2)^5x = sqrt(2)/4. Next, we can take the logarithm of both sides with base 2 to get 5x = log2(sqrt(2)/4). Finally, we can solve for x by dividing both sides by 5 to get x = (1/5) * log2(sqrt(2)/4).

2. Can you explain the concept of an exponential equation in simpler terms?

An exponential equation is one in which the variable is in the exponent. This means that the value of the variable is being multiplied by itself a certain number of times. For example, in the equation 2^x = 16, the variable x is being multiplied by itself 4 times (2^4 = 16).

3. How do I know when to use logarithms to solve an exponential equation?

You should use logarithms when the variable is in the exponent and you need to solve for the variable. In other words, when you have an equation of the form a^x = b, where a is a constant and x is the variable, you can use logarithms to solve for x.

4. Why do we need to take the logarithm of both sides when solving an exponential equation?

Taking the logarithm of both sides allows us to bring the variable out of the exponent and solve for it. The logarithm function is the inverse of the exponential function, so it allows us to "undo" the exponentiation and solve for the variable.

5. Can you give an example of a real-life situation in which an exponential equation would be used?

An example of a real-life situation in which an exponential equation would be used is in modeling population growth. As a population grows, it follows an exponential pattern - the larger the population, the faster it grows. This can be represented by the equation P(t) = P(0) * e^(rt), where P(0) is the initial population, r is the growth rate, and t is the time. By solving for t, we can predict when the population will reach a certain size.

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