Unknown in logarithm base $122312/100000 = (1+t)^5$ I am new to exponents and logarithms, and have been stuck with this for a quite long time.
The problem is:
$$\frac{122312}{100000}=(1+t)^5$$
or
$$\log_{1+t}\frac{122312}{100000}=5$$
I have no idea, how to solve this. Hints or tips would be very very welcomed.
Thank You (a lot) in advance!(also apologies for my poor english)
 A: Why are you taking the logarithm? I think you should compute the fifth root of $\frac{122312}{100000}$ and then subtract $1$ from it to get $t$.
You can use the log table to compute the fifth root of $\frac{122312}{100000}$. (Assuming you know how to use it) Let
$$n={\left(\frac{122312}{100000}\right)}^{\frac{1}{5}}$$
then take log on both sides to get
$$\log_{10}(n)={\frac{1}{5}}\log_{10}(1.22312)$$
First compute $\log_{10}(n)$ and then using the antilog tables, compute $n$. Then, since $t+1=n$, $t=n-1$
A: The first equation says that
$$1+t=\sqrt[5]{\frac{122312}{100000}}=\sqrt[5]{1.22312}=1.22312^{1/5}\;;$$
if you’re required only to get a good numerical approximation to $t$, just use any convenient calculator to get
$$t=1.22312^{1/5}-1\approx0.0411033$$
(or however precise an approximation is wanted).
Any exact expression for $t$ is going to be fairly ugly, and without more context it’s not clear even what sort of expression would be wanted.
A: $$\frac{122312}{100000}=(1+t)^5$$
$$1.22312=(1+t)^5$$
$$1+t=\sqrt[5]{1,22312}$$
$$t=\sqrt[5]{1,22312}-1$$
A: If you really want to go through logarithms, you can do as follows:
$$\frac{122312}{100000}=(1+t)^5$$
$$\log_{1+t}\left(\frac{122312}{100000}\right) = 5$$
$$\frac{\ln\left(\frac{122312}{100000}\right)}{\ln(1+t)} = 5$$
$$\ln(t+1) = \frac{\ln\left(\frac{122312}{100000}\right)}{5}$$
$$t = e^\frac{\ln\left(\frac{122312}{100000}\right)}{5} - 1$$
$$t = \left(\frac{122312}{100000}\right)^\frac{1}{5} - 1$$
which is exactly the solution you get by usual means.
