Logs with diff. Base: log(base5)x - 4 = log(base7)x

[D!ddy]

Hi I'm really stopped on this problem it says to solve for x

log(base 5)x-4 = log(base 7)x And round answer to 4 decimal places

Instead of doing algebraically I used my graphing calculator to find the answer which is x is approx. 11.5842, but I don't know how to get there algebraically, so if anyone could help I`d really appreciate it.

Diddy

 

[stapel]

...solve for x: log(base 5)x-4 = log(base 7)x And round answer to 4 decimal places

Your formatting is ambiguous. I will guess that you mean the following:

. . . . .log[sub:q0pnfs4r]5[/sub:q0pnfs4r](x) - 4 = log[sub:q0pnfs4r]7[/sub:q0pnfs4r](x)

Probably the simplest solution method would be to apply the Change of Base formula that you've memorized, using whatever base you like. Using the natural log, followed by some rearrangement, you should get:

. . . . .ln(x) = [ 4 ln(5) ln(7) ] / [ ln(7/5) ]

Raise either side as a power on "e", and find the decimal approximation in your calculator. :wink:

Note: I could not obtain "x = 11.58 (approx)" as a solution for either log[sub:q0pnfs4r]5[/sub:q0pnfs4r](x) - 4 or log[sub:q0pnfs4r]5[/sub:q0pnfs4r](x - 4). :shock:

When you reply, please show all of your work and reasoning. Thank you!

Eliz.

 

[D!ddy]

. . . . .ln(x) = [ 4 ln(5) ln(7) ] / [ ln(7/5) ]

Raise either side as a power on "e", and find the decimal approximation in your calculator. :wink:

Well for this question, I haven't been taught the "Ln" function in my calculator, yet or the power "e", but I have learned the Change of Base formula
so the work I did is...
log5(x-4) = log7(x)
log5(x-4) = log5(x) / log5(7)
log5(x-4) * log5(7) = log5(x)
log5(x-4)+(7) = log5(x)
log5(x+3) = log5(x)
x+3 = x

As you can see I got stuck at the end because I have to solve for x, I changed the base to 5 so it would seem easier but it seems not.
But I did a check of x = 11.5842 and both sides are equal when I used that, so there you go...sorry if it doesn't help

 

[o_O]

How'd you get from log[sub:1p4y580o]5[/sub:1p4y580o]7 equal to 7 from the 3rd to 4th line?

And how did you get from log[sub:1p4y580o]5[/sub:1p4y580o](x - 4) + 7 to log[sub:1p4y580o]5[/sub:1p4y580o](x + 3) on the 4th to 5th line? You can't add a number to something inside the logarithm ...

 

[D!ddy]

How'd you get from log[sub:doysd1k8]5[/sub:doysd1k8]7 equal to 7 from the 3rd to 4th line?

And how did you get from log[sub:doysd1k8]5[/sub:doysd1k8](x - 4) + 7 to log[sub:doysd1k8]5[/sub:doysd1k8](x + 3) on the 4th to 5th line? You can't add a number to something inside the logarithm ...

OMG thanks i just noticed I got the Product Rule upside down
but wait since they're all the same base i can remove the logs now right?

 

[stapel]

...since they're all the same base i can remove the logs now right?

Please reply showing your work and reasoning, so we can "see" what you mean.

Thank you!

Eliz.

 

[D!ddy]

...since they're all the same base i can remove the logs now right?

Please reply showing your work and reasoning, so we can "see" what you mean.

Thank you!

Eliz.

log[sub:2v2umean]5[/sub:2v2umean](x-4) = log[sub:2v2umean]7[/sub:2v2umean](x)

change base to 5
log[sub:2v2umean]5[/sub:2v2umean](x-4) = log[sub:2v2umean]5[/sub:2v2umean](x) / log[sub:2v2umean]5[/sub:2v2umean](7)
log[sub:2v2umean]5[/sub:2v2umean](x-4) * log[sub:2v2umean]5[/sub:2v2umean](7) = log[sub:2v2umean]5[/sub:2v2umean](x)

this is what i meant by "removing" the logs...
(x-4) * (7) = (x)
7x - 28 = x
6x = 28
x = 14/3


I know it might sound stupid but I was wondering since log is the inverse of exponent functions..in my graphing calculator when i graphed y[sub:2v2umean]1[/sub:2v2umean] = log[sub:2v2umean]5[/sub:2v2umean](x-4) and y[sub:2v2umean]2[/sub:2v2umean] = log[sub:2v2umean]7[/sub:2v2umean](x) and used the INTERSECTION function do i use the x value of the intersection point or the y value? because the y value of the intersection point is the same number I found on the when I plug in 11.5842.. when I checked both sides of the original equation, using the L.S and R.S check thing...I hope this helps I've been stuck on this problem for a week now. lol but I hate it when I get stuck on a problem

 

[Deleted member 4993]

[quote="D!ddy":2log3r0y]...since they're all the same base i can remove the logs now right?

Please reply showing your work and reasoning, so we can "see" what you mean.

Thank you!

Eliz.

log[sub:2log3r0y]5[/sub:2log3r0y](x-4) = log[sub:2log3r0y]7[/sub:2log3r0y](x)

change base to 5
log[sub:2log3r0y]5[/sub:2log3r0y](x-4) = log[sub:2log3r0y]5[/sub:2log3r0y](x) / log[sub:2log3r0y]5[/sub:2log3r0y](7)
log[sub:2log3r0y]5[/sub:2log3r0y](x-4) * log[sub:2log3r0y]5[/sub:2log3r0y](7) = log[sub:2log3r0y]5[/sub:2log3r0y](x)

this is what i meant by "removing" the logs...
(x-4) * (7) = (x) <--- No you cannot do that.

However

log[sub:2log3r0y]5[/sub:2log3r0y](x-4) * log[sub:2log3r0y]5[/sub:2log3r0y](7) = log[sub:2log3r0y]5[/sub:2log3r0y](x)

log[sub:2log3r0y]5[/sub:2log3r0y](x-4) * 1.209061955 = log[sub:2log3r0y]5[/sub:2log3r0y](x)

log[sub:2log3r0y]5[/sub:2log3r0y]{(x-4) [sup:2log3r0y]1.209061955[/sup:2log3r0y]} = log[sub:2log3r0y]5[/sub:2log3r0y](x)

Now you can remove logs to get:

(x-4) [sup:2log3r0y]1.209061955[/sup:2log3r0y] = x

This is a horribly non-linear equation - not amenable to simple algebraic manipulations.


7x - 28 = x
6x = 28
x = 14/3
[/color]

I know it might sound stupid but I was wondering since log is the inverse of exponent functions..in my graphing calculator when i graphed y[sub:2log3r0y]1[/sub:2log3r0y] = log[sub:2log3r0y]5[/sub:2log3r0y](x-4) and y[sub:2log3r0y]2[/sub:2log3r0y] = log[sub:2log3r0y]7[/sub:2log3r0y](x) and used the INTERSECTION function do i use the x value of the intersection point or the y value? because the y value of the intersection point is the same number I found on the when I plug in 11.5842.. when I checked both sides of the original equation, using the L.S and R.S check thing...I hope this helps I've been stuck on this problem for a week now. lol but I hate it when I get stuck on a problem[/quote:2log3r0y]