Logs and algebra question

[walker8476]

I'm studying logs at the moment and having trouble with this question which has algebra in it as well:

Transpose this equation to make y the subject

10^y = 1 + x

The answer is y =log (1 + x)


My attempt:

10^y = 1 + x

log10^y = log (1 + x)

ylog10 = log (1 + x)

y = log (1 + x) / log10

 

[stapel]

I'm studying logs at the moment and having trouble with this question which has algebra in it as well:

Transpose this equation to make y the subject

10^y = 1 + x

The answer is y =log (1 + x)

You can find a definition-based shortcut here.

My attempt:

10^y = 1 + x

log10^y = log (1 + x)

ylog10 = log (1 + x)

y = log (1 + x) / log10

Thank you for showing your work so nicely!

Now, apply the definition of logarithms. What power would be required to raise the base of 10 to end up with a value of 10? That is, 10^x = 10 for what value of x? This is the numerical value of log₁₀(10). Plug this in, and simplify.

 

[Ishuda]

I'm studying logs at the moment and having trouble with this question which has algebra in it as well:

Transpose this equation to make y the subject

10^y = 1 + x

The answer is y =log (1 + x)


My attempt:

10^y = 1 + x

log10^y = log (1 + x)

ylog10 = log (1 + x)

y = log (1 + x) / log10

Your answer is correct no matter what base you use for the logarithm. That is
\(\displaystyle y = \frac{log_b(1+x)}{log_b(10)}\)
for any base you want to use.

However, saying the same thing stapel did in a different way, apparently they want the answer for a particular base. Now
log_b(b) = 1
for any base b. So, if
log(10) = 1
what is the base?

 

[walker8476]

You can find a definition-based shortcut here.


Now, apply the definition of logarithms. What power would be required to raise the base of 10 to end up with a value of 10? That is, 10^x = 10 for what value of x? This is the numerical value of log₁₀(10). Plug this in, and simplify.

So log10 = 1. So ......

y = log (1 + x) / 1

y = log (1+ x) ???

 

[stapel]

So log10 = 1. So ......

y = log (1 + x) / 1

y = log (1+ x) ???

What is your question?

 

[walker8476]

What is your question?

Is it correct?

 

[stapel]

Is it correct?

Does it match the answer you'd been given?

 

[walker8476]

Does it match the answer you'd been given?

You could just say "Yes that's correct"

 

[HallsofIvy]

So you believe you have to be told the answer and are not capable of checking for yourself?