solving exponential equation: 3^(-8x) = 4^(x - 5)

amanda_burnett

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3[sup:1ynohqhk]-8x[/sup:1ynohqhk] = 4[sup:1ynohqhk](x-5)[/sup:1ynohqhk]

Solve the equation for x
Write your answer as an expression involving base-10 logarithms.
 
Re: solving an exponential equation

the first step I thought I could do would be

-8x log[sub:3vxos02t]10[/sub:3vxos02t] 3 = (x-5) log[sub:3vxos02t]10[/sub:3vxos02t] 4

but after that I don't know were to go
 
Re: solving an exponential equation

amanda_burnett said:
the first step I thought I could do would be

-8x log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 3 = (x-5) log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 4

but after that I don't know were to go

Then isolate 'x'

-8x log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 3 = (x-5) log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 4

-8 log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 3 / log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 4 = (x-5)/x

-8 log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 3 / log[sub:1v1nvr6p]10[/sub:1v1nvr6p] 4 = 1 - 5/x

and contimue....
 
Re: solving an exponential equation

Then isolate 'x'

-8x log10 3 = (x-5) log10 4

-8 log10 3 / log10 4 = (x-5)/x

-8 log10 3 / log10 4 = 1 - 5/x


How did you put in a one instead of a x for that last step?

then it would be
-8 log[sub:3npgrju8]10[/sub:3npgrju8]3 = -4/x
log[sub:3npgrju8]10[/sub:3npgrju8]4
 
Re: solving an exponential equation

would the next step be
-6 = -4 / x

because log[sub:20yvylrd]10[/sub:20yvylrd] = 1
 
Re: solving an exponential equation

is the final answer x = 2/3

but it says to write the answer as an expression involving base-10 logarithms
 
Re: solving an exponential equation

amanda_burnett said:
Then isolate 'x'

-8x log10 3 = (x-5) log10 4

-8 log10 3 / log10 4 = (x-5)/x

-8 log10 3 / log10 4 = (x/x) - (5/x)

5/x = 1 + 8 log10 3 / log10 4

Now continue.....

How did you put in a one instead of a x for that last step?

then it would be
-8 log[sub:3d2ldndj]10[/sub:3d2ldndj]3 = -4/x <<<< NO! You are for getting PEMDAS
log[sub:3d2ldndj]10[/sub:3d2ldndj]4
 
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