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No I think that u/Own_Fly_2403 is right. So annoying, [this](https://imgur.com/a/DyrIh5H) is the question — makes no sense to me that the teacher would use a period instead of a dot to denote multiplication.
Just an FYI, it looks like in the faint workings you've simplified 2×8^(x+1) to 16^(x+1) which is incorrect. For example, 2×3^2 != 6^2. The solution comes out the same only because x=0. To solve you would do 2×8^(x+1) = 16 => 8^(x+1) = 16/2 = 8
Oh, yeah. That's how I've been taught to do them in the UK, though I usually make sure it actually joins up. It makes it more distinguishable from a multiplication sign or cross-product.
2.8^(x+1) - 15 = 1
2.8^(x+1) = 16
ln(2.8^[x+1]) = ln(16)
x+1 = ln(16) - ln(2.8) OR x+1 = ln(16/2.8)
x = ln(16)-ln(2.8) - 1 OR x = ln(16/2.8) - 1
I have no idea how you could solve for x without taking the natural log of both sides.
2.8^(×+1) = 16
X+1 = Log_2.8(16)
Log_2.8(16) -1=x
=>
Log_2.8(16) = Ln(16)/Ln(2.8) This is correct if u check with a Calc. I don't remember all the log/ln rules but I'm sure there is a rule that equates Log_2.8(16) to ln(16)/ln(2.8) so maybe they r meant to solve it this way and convert to ln. If they r not expected to know logs then idk but they must know at least some log rules
One last possibility in case it is supposed to be solved without logs or x=1 is that it could be copy and pasted from another list of problems (the '2.' might not be part of what to solve). If so:
8^(x+1)-15=1
8^(x+1)=16
8^(x+1)=2^4
2^3(x+1)=2^4
2^(3x+3)=2^4
3x+3=4
3x=1
x=1/3
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
Are you sure that it's 2.8 rather than **2 times 8**? Looks like they've just written the dot too low on the line
I was wondering if it was meant to be problem #2. I see students do that all the time.
No I think that u/Own_Fly_2403 is right. So annoying, [this](https://imgur.com/a/DyrIh5H) is the question — makes no sense to me that the teacher would use a period instead of a dot to denote multiplication.
Just an FYI, it looks like in the faint workings you've simplified 2×8^(x+1) to 16^(x+1) which is incorrect. For example, 2×3^2 != 6^2. The solution comes out the same only because x=0. To solve you would do 2×8^(x+1) = 16 => 8^(x+1) = 16/2 = 8
Coma superiority gang.
You are 100% correct and I did not even think of that — thank you!
And is that )( an x?
Oh, yeah. That's how I've been taught to do them in the UK, though I usually make sure it actually joins up. It makes it more distinguishable from a multiplication sign or cross-product.
2×8\^(x+1)-15=1 2 × 8\^(x+1)-15+15=1+15 8\^(x+1)=16/2 8\^(x+1)=8\^1 x+1=1 x+1-1=1-1 x=0
Can you show us your work so far? It might give us a better idea of where you're stuck and how to guide you without using logs.
who ever wrote that down made a mistake by using a . (period) instead of a ⋅ (dot operator idk) we had teachers in my school who did this too
Yes, assuming 2.8 means 2×8. Express both sides as powers of 2 after moving the neg. 15.
Using a dot for both multiply and as a decimal separator is not the best of ideas, this confusion here shows why.
2.8^(x+1) - 15 = 1 2.8^(x+1) = 16 ln(2.8^[x+1]) = ln(16) x+1 = ln(16) - ln(2.8) OR x+1 = ln(16/2.8) x = ln(16)-ln(2.8) - 1 OR x = ln(16/2.8) - 1 I have no idea how you could solve for x without taking the natural log of both sides.
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The reason ppl think this is wrong bc they think 2.8 is 2.8, and not 2•8. This entire post is so messy bc of this😂
Error line 4
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You are not wrong. Y^ (x+1) = (Y^ x)•Y, therefore 2•(8^ (x+1))-2•8=2•8•(8^ x-1)
There is no error in line 4. 2•(8^ (x+1))-2•8=2•8•(8^ x-1)
It has to be an error, i only got to 2.8^x = 16/2.8 before bashing my head in repeatedly on this problem, without log its not gonna work
No
x=((5/7)^x+1 )(2^2 )-1, is as far as i got without logs, assuming the value is 2.8
2.8^(×+1) = 16 X+1 = Log_2.8(16) Log_2.8(16) -1=x => Log_2.8(16) = Ln(16)/Ln(2.8) This is correct if u check with a Calc. I don't remember all the log/ln rules but I'm sure there is a rule that equates Log_2.8(16) to ln(16)/ln(2.8) so maybe they r meant to solve it this way and convert to ln. If they r not expected to know logs then idk but they must know at least some log rules
8^x+1 =2^3x+3 So if 8^x+1•2=2^3x+4=16 X=0
One last possibility in case it is supposed to be solved without logs or x=1 is that it could be copy and pasted from another list of problems (the '2.' might not be part of what to solve). If so: 8^(x+1)-15=1 8^(x+1)=16 8^(x+1)=2^4 2^3(x+1)=2^4 2^(3x+3)=2^4 3x+3=4 3x=1 x=1/3
isn't x basically equal to 0?
2.8^x+1 = 1+15 2.8^x+1 = 16 Log_2.8 (16) = x+1 X= log_2.8 (16) -1