Logarithms and exponential functions Watch

Shas72
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5^(x^2)<2^x solve giving your ans in terms of log
I got the ans x>log2/log 5, but the other ans is x<0. I dont understand how is x<0.
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dextrous63
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Think you made the classic mistake of dividing by x instead of taking everything to one side and factorising.

Edit. Are you sure the inequality in the original question is the right way round?
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Shas72
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(Original post by dextrous63)
Think you made the classic mistake of dividing by x instead of taking everything to one side and factorising.
No I did factorising .so 5^2x-2^x>0 and then applying log I get x>0 but the ans is x<0
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dextrous63
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(Original post by Shas72)
No I did factorising .so 5^2x-2^x>0 and then applying log I get x>0 but the ans is x<0
Er, how did you apply logs after shifting everything over? By the looks of it, you seem to have made a common mistake:

a + b = c does NOT mean log a + log b = log c
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Shas72
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(Original post by dextrous63)
Er, how did you apply logs after shifting everything over? By the looks of it, you seem to have made a common mistake:

a + b = c does NOT mean log a + log b = log c
I dont understand
Can you pls explain a bit more
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Shas72
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So I did x^2log5-xlog2>0. Factorise you get x>0 and x> log2/log5
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dextrous63
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(Original post by Shas72)
So I did x^2log5-xlog2>0. Factorise you get x>0 and x> log2/log5
OK. Your other post didn't make sense, hence my question. Back to the task.

Sketch the quadratic curve y=log5 x^2 - log2 x, marking the roots on the x-axis.
When is the curve above the x-axis, and thus positive?
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Shas72
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(Original post by dextrous63)
OK. Your other post didn't make sense, hence my question. Back to the task.

Sketch the quadratic curve y=log5 x^2 - log2 x, marking the roots on the x-axis.
When is the curve above the x-axis, and thus positive?
How do you do that? So you have to calculate the values of log 5 and log 2
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dextrous63
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(Original post by Shas72)
How do you do that? So you have to calculate the values of log 5 and log 2
No. You know it's a quadratic, and factorising will give x(log5 x - log2), so the roots are 0 and log2/log5.
Sketch the curve, marking these two point on the x-axis.
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Shas72
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(Original post by dextrous63)
No. You know it's a quadratic, and factorising will give x(log5 x - log2), so the roots are 0 and log2/log5.
Sketch the curve, marking these two point on the x-axis.
So I marked 0 and 0.4 on x axis
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dextrous63
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(Original post by Shas72)
So I marked 0 and 0.4 on x axis
Sketch the curve. When does it go above the x-axis?
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Shas72
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(Original post by dextrous63)
No. You know it's a quadratic, and factorising will give x(log5 x - log2), so the roots are 0 and log2/log5.
Sketch the curve, marking these two point on the x-axis.
I really dont

(Original post by dextrous63)
Sketch the curve. When does it go above the x-axis?
Iam sorry but I dont understand.
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dextrous63
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(Original post by Shas72)
I really dont


Iam sorry but I dont understand.
What does a quadratic curve look like? Where does it pass through the x-axis?
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Shas72
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(Original post by dextrous63)
What does a quadratic curve look like? Where does it pass through the x-axis?
Quadratic curve is a parabola . It passes x axis when x=0 and y=0
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dextrous63
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(Original post by Shas72)
Quadratic curve is a parabola . It passes x axis when x=0 and y=0
I think there are some holes in your understanding which need dealing with.

With that in mind, either download the Desmos app on your phone, or go to geogebra on your laptop/phone/tablet and get it to draw the graph y=log5 x^2 - log2 x (you might find it easier to use decimal values for the logs) and look for where the graph is positive (above the x-axis).
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Shas72
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(Original post by dextrous63)
I think there are some holes in your understanding which need dealing with.

With that in mind, either download the Desmos app on your phone, or go to geogebra on your laptop/phone/tablet and get it to draw the graph y=log5 x^2 - log2 x (you might find it easier to use decimal values for the logs) and look for where the graph is positive (above the x-axis).
Thanks will do that
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Shas72
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(Original post by dextrous63)
I think there are some holes in your understanding which need dealing with.

With that in mind, either download the Desmos app on your phone, or go to geogebra on your laptop/phone/tablet and get it to draw the graph y=log5 x^2 - log2 x (you might find it easier to use decimal values for the logs) and look for where the graph is positive (above the x-axis).
I think I understood what you meant
Its above x axis only when x<0
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dextrous63
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And after 0.4
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