# How to solve this exponential equation?

Homework Statement:
Solve the equation: (2x+5)^(x+1)=729
Relevant Equations:
(2x+5)^(x+1)=729
Homework Statement: Solve the equation: (2x+5)^(x+1)=729
Homework Equations: (2x+5)^(x+1)=729

Here's my work:
(2x+5)^(x+1)=729
x+1=ln(729)/ln(2x+5)
x+1=ln(729/(2x+5))
e^(x+1)=e^ln(729/(2x+5))
e^(x+1)=729/(2x+5)
From here, I don't know what to do. All I know is that the answer is x=2. But I don't know how to get there. Can someone please help me?

fresh_42
Mentor
It is always a good idea to look at what you have!

On the LHS are ##x+1## equal factors. Which factors do you have on the RHS?

HallsofIvy
SammyS
Staff Emeritus
Homework Helper
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Homework Statement: Solve the equation: (2x+5)^(x+1)=729
Homework Equations: (2x+5)^(x+1)=729

Here's my work:
(2x+5)^(x+1)=729
x+1=ln(729)/ln(2x+5)
x+1=ln(729/(2x+5))
e^(x+1)=e^ln(729/(2x+5))
e^(x+1)=729/(2x+5)
From here, I don't know what to do. All I know is that the answer is x=2. But I don't know how to get there. Can someone please help me?
You do have an error in your workings.
ln(729)/ln(2x+5) is not equal to ln(729/(2x+5)). In general, ##\dfrac{\ln(a)}{\ln(b)} \ne \ln \dfrac a b ## .

Furthermore, I doubt that your approach is likely to give a solution.

To expand slightly on fresh_42's hint: Can you write 729 as some number raised to some other number?

sysprog
haruspex
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2020 Award
I know is that the answer is x=2. But I don't know how to get there.
Since it is easy enough to check that x=2 is an answer, I assume you are asking how to arrive at that by algebraic steps. The short answer is that you cannot.

To solve a question like that in general, without being given a solution, you need to be a bit creative. First questions should be whether the solution is likely to exist and be unique. In the present case you can check this by considering the ranges over which the function on the left is monotonic and for which of these the result encompasses 729.
Next, you can "binary chop", subdividing ranges to narrow down where solutions should lie.

sysprog
##9^3=3^6=3^{2^3}=729##

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Not entirely correct.

##3^4=81##, not ##729##.

##9^3=(3\times3)^3=3^6##
You're right -- my mistake -- I edited my post to correct it.

SammyS
Staff Emeritus
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##9^3=3^6=3^{2^3}=729##
New error ##3^6## is correct. ##3^{2^3}## is not correct.

##3^{2^3} =3^8 =6561## ##\ \ne 729##

Last edited:
New error ##3^6## is correct. ##3^{2^3}## is not correct.

##3^{2^3} =19683## ##\ \ne 729##
I guess that's a 'forgetting the rule' error -- I meant ##{(3^2)}^3##; too late for me to re-edit, so here's the second corrected version: ##9^3=3^6={(3^2)}^3=729##

fresh_42
Mentor
A reminder @all:

Please do not solve the problem for the OP! Remember that we want to teach, not solve trivial problems. I suppose everybody answering here saw the solution, so there is no need to tell the world the obvious.

A reminder @all:

Please do not solve the problem for the OP! Remember that we want to teach, not solve trivial problems. I suppose everybody answering here saw the solution, so there is no need to tell the world the obvious.
Sorry -- didn't know I was overdoing it -- I'll be more careful about that.

##3^{2^3} =3^8 =6561## ##\ \ne 729##

I saw ##19683## in your post, and hit reply to ask how that came about, but when I saw the ##\LaTeX## it said 6561 -- it looked like this: #3^{2^3} =3^8 =6561## ##\ \ne 729#-- I suppose maybe more braces were in order somewhere.

SammyS
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##3^{2^3} =3^8 =6561## ##\ \ne 729##

I saw ##19683## in your post, and hit reply to ask how that came about, but when I saw the ##\LaTeX## it said 6561 -- it looked like this: #3^{2^3} =3^8 =6561## ##\ \ne 729#-- I suppose maybe more braces were in order somewhere.
You must have hit the reply button before I edited and re-submitted my post. What I had initially calculated was ##3^{3^2} ## (by mistake), which is ##19683##.
My edited post was submitted about 13 minutes after the initial post.

Last edited:
sysprog