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Logarithms and exponential functions

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
Reply 1
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.

So instead of having y in your working what do you have instead?
Reply 2
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.

So instead of having y in your working what do you have instead?

Can you find the rule for the exponential function by writing y in terms of x ?
Reply 3
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
Reply 4
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
Reply 5
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.

So instead of having y in your working what do you have instead?

I hadn't paid attention on the graph that's plotted between log base 10 and x. I understood your point. Thanks a lot
Reply 6
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
Reply 7
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.
Reply 8
Original post by Shas72
Thanks a lot for explaining in detail.

No worries!

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