Logarithms and exponential functions

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.

Reply 1

T!gger34

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?

(edited 4 years ago)

Reply 2

Shas72

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

Reply 3

T!gger34

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

Reply 4

Shas72

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

Reply 5

Shas72

So I did x^2log5-xlog2>0. Factorise you get x>0 and x> log2/log5

Reply 6

T!gger34

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?

Reply 7

Shas72

Original post by dextrous63

How do you do that? So you have to calculate the values of log 5 and log 2

Reply 8

T!gger34

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.

Reply 9

Shas72

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

Reply 10

T!gger34

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?

Reply 11

Shas72

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.

Reply 12

T!gger34

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?

Reply 13

Shas72

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

Reply 14

T!gger34

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).