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)
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 $t+1=n$, $t=n-1$
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.
$$\frac{122312}{100000}=(1+t)^5$$ $$1.22312=(1+t)^5$$ $$1+t=\sqrt[5]{1,22312}$$ $$t=\sqrt[5]{1,22312}-1$$
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.
