# Logarithms and exponential functions

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#1
Variables x and y are related so that when log (base 10) y is plotted on the vertical axis and x is plotted on the horizontal axis , a straight line graph passing thro the points (2,5) and (6,11) is obtained.
1) express log (base10) y in terms of x
So when I do this I get y=3x/2 +2. So i apply log base 10 on both sides. I get log (base 10) y= log (base10) (3x/2+2). But the ans is log base 10 y= 3x/2+2. I dont understand
0
1 month ago
#2
(Original post by Shas72)
Variables x and y are related so that when log (base 10) y is plotted on the vertical axis and x is plotted on the horizontal axis , a straight line graph passing thro the points (2,5) and (6,11) is obtained.
1) express log (base10) y in terms of x
So when I do this I get y=3x/2 +2. So i apply log base 10 on both sides. I get log (base 10) y= log (base10) (3x/2+2). But the ans is log base 10 y= 3x/2+2. I dont understand
What you need to remember is that the points you are given lie on a straight line produced by graphing log₁₀y versus x not y versus x as you seem to think.

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1 month ago
#3
(Original post by hoosie)
What you need to remember is that the points you are given lie on a straight line produced by graphing log₁₀y versus x not y versus x as you seem to think.

Can you find the rule for the exponential function by writing y in terms of x ?
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#4
(Original post by hoosie)
Can you find the rule for the exponential function by writing y in terms of x ?
Is it y= 10^3x/2+2
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#5
(Original post by hoosie)
Can you find the rule for the exponential function by writing y in terms of x ?
I got it thanks a lot
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#6
(Original post by hoosie)
What you need to remember is that the points you are given lie on a straight line produced by graphing log₁₀y versus x not y versus x as you seem to think.

I hadn't paid attention on the graph that's plotted between log base 10 and x. I understood your point. Thanks a lot
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1 month ago
#7
(Original post by Shas72)
Is it y= 10^3x/2+2
log₁₀y = 3x/2 + 2
log₁₀y = 3x/2 + log₁₀100
log₁₀y - log₁₀100 = 3x/2
log₁₀(y/100) = 3x/2
y/100 = 10¹·⁵ˣ
y = 100(10¹·⁵ˣ )
Check: when x = 2,
y = 100(10³)
⇒ y = 100 000
log₁₀100 000 = 5
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#8
(Original post by hoosie)
log₁₀y = 3x/2 + 2
log₁₀y = 3x/2 + log₁₀100
log₁₀y - log₁₀100 = 3x/2
log₁₀(y/100) = 3x/2
y/100 = 10¹·⁵ˣ
y = 100(10¹·⁵ˣ )
Check: when x = 2,
y = 100(10³)
⇒ y = 100 000
log₁₀100 000 = 5
Thanks a lot for explaining in detail.
0
1 month ago
#9
(Original post by Shas72)
Thanks a lot for explaining in detail.
No worries!
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