Find the exact solution to the equation. 100(1/5)^x/4=4 1/2 8 2 9
Could you possibly take a screen shot of this problem and then upload that screen shot to OpenStudy? I'm having trouble deciphering what you've just typed in.
Can you identify "base," "number," "exponent" here?
\[100(1/5)^{x/4}=4\]
that's so much better! What is the base of this exponential function? What is the goal of this problem?
1/5 is base
Right, it is. \[100(1/5)^{x/4}=4\]
Here you see 100 as a multiplier and (1/5)^ (x/4) as a power function. Which shall we move to the right side as part of the process of soliving for x?
100
Very good choice. Pls divide both sides of this equation by 100 and type out the results.
1/5^x/4=4/100
Must use parentheses around the base "1/5" to reduce or eliminate any ambiguity in what is meant. This is quite important. So, you get \[(1/5)^\frac{ x }{ 4 }=0.04\]
What's oour goal here? What's a logical first step towards achieving that goal?
goal is to get x by its self. FIrst step is to get 1/5 to the other side
The situation here is different in that (x/4) is an EXPONENT, not a multiplier. Remember that we're talking about bases and powers here. Have to use that approach to solve for x/4 and then for x alone.
I think you'll find the problem easier to solve if you'd take the 4th power of both sides of the equation. Mind doing that?
Hint:\[(x ^{1/4})^4=x\]
1/5^x=.044^4
Again, you must enclose that fraction 1/5 inside parentheses. Also, on the right we have 0.04^4, not 0.044^4. Please try again.
(1/5)^x=(.04)^4
Good. Now, you must somehow isolate x. Remember, we've been discussing expo functions and power functions. What is the base of the power function shown here? How would you use logs to solve for x?
idk
Remember the following? \[\log_{a}a^x=x? \]
Is that of any help?
no
Naveen, you will definitely have to learn, understand and apply that principle as you study exponential and log functions. I strongly recommend that you write down the several formulas I've shared with you here and keep that list for future reference.
What is \[\log_{2}2^1? \]
ok i will but for now can you please just tell me the answer
I'm sorry, I don't operate that way, and for me to give you the answer without your involvement would violate OpenStudy's Code of Conduct. We are close to solving the problem at hand.
As I explained earlier, the log function "undoes" the expo function, and the expo function "undoes" the log function. \[2^{\log_{2}x }=x\]
so the 2 in front cancels out
... is one eexample of that.\[2^{\log_{2}x }\]
...is another exampler, and, like the first one, comes out to "x" in the end.
We need to solve (1/5)^x=(.04)^4 for x. One way to do this would be to take the base 10 log of both sides, as shown below:
\[\log (\frac{ 1 }{ 5 })^x=\log 0.04^4\]
Are you familiar enough with the rules of logs to be able to simplify the left and right sides of this equation (separately)?
not confident
I can help and support you, but in the end you do have to learn, study and apply the concepts I've discussed with you here. \[\log (\frac{ 1 }{ 5 })^x=\log 0.04^4\] is to be simplified.
There is a rule of logs that goes like this:\[\log a^b=b*\log a\]
and so\[\log (1/5)^x=x*\log (1/5)\]
and, for the right side, \[\log 0.04^4=4*\log 0.04 \]
Can you simplify log (1/5)? The pertinent rule is \[\log \frac{ a }{ b }=\log a-\log b\]
I just did it answer to the equation is 8
Naveen? How do you normally learn new material, and from where do you learn it? If most or all of what I'm typing to you here is new and unfamiliar to you, this work will be very hard. I'm glad to help, but want to ensure that you realize that a lot of background material is needed here to enable you to solve the problems at hand.
I just have to do many problems until it hits me.
Hope you are writing down the formulas and background info that I'm sharing with you. \[\log (1/5)^x=x*\log (1/5).becomes.x(\log 1 - \log 5)\]
if we follow the rule I just gave you. log 1 is zero and log 5 can be left as is. Therefore, the left side of our equation\[\log (\frac{ 1 }{ 5 })^x=\log 0.04^4\]
becomes x(0-log 5). We now have to tackle the right side.
The rule for that exponentiation I have already presented, but I'll present it again:\[\log a^b=b*\log a\] so our log 0.04^4 becomes\[4 \log 0.04=4(\log 4+\log 10^{-2}).\]
Do you remember how to evaluate this last result?
OpenStudy reports that y ou're looking at someone else's math problem. You and I have a lot of time invested in discussing this post of yours, and I really hope we can finish the work done. What about you?
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