Exponential function (t)

Question

I got the function $8.513 \times 1.00531^{\Large t} = 10$. The task is to solve $t$. The correct answer is $t = 31$. How do I get there ?.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &8.513\times 1.00531^{t}=10\ \imp\ 1.00531^{t}={10 \over 8.513}\ \imp\ \ln\pars{1.00531^{t}} = \ln\pars{10 \over 8.513} \\[3mm]&\imp t\ln\pars{1.00531} = \ln\pars{10 \over 8.513}\ \imp\ \color{#66f}{\large t = {\ln\pars{10/8.513} \over \ln\pars{1.00531}}} \approx {\tt 30.3988} \end{align}

Answer 2

Hint: Use the fact that $\log(ab^c)=\log a + c\log b$

Answer 3

Hint: $$8.513 \cdot 1.00531^t=10 \iff 1.00531^t=\frac{10}{8.513} \iff \ln[1.00531^t]=\ln\left[\frac{10}{8.513}\right] $$ $$\iff t\ln[1.00531]=\ln\left[\frac{10}{8.513}\right] \iff t= \quad?$$

Answer 4

• Isolate $ 1.00531^{\large t}$ on the left side of the equation.

• Take the $\ln$ of each side of that equation.

• And use the fact that $\ln a^b = b \ln a$.

• Solve for $t$ as you would any first degree polynomial.