\therefore x = 2 &\text{ or } x = 3 To see what happened to the solution \(x=1\), we substitute it into our original equation to obtain \(\log_{117}(-2) = \log_{117}(-2)\). Graphing \(f(x) = 1 + \frac{2\ln(x+1)}{\ln(4)}\) and \(g(x) = \frac{2 \ln(x)}{\ln(2)}\), we see the graphs intersect only at \(x = 1 + \sqrt{3} \approx 2.732\). \end{align*}, \begin{align*} This is commonly referred to as taking the logarithm of both sides. On the left side, we x - 4 + 4, which equals x, and on the right side, we get 5 + 4, which equals 9. \frac{27^x - 1}{9^x + 3^x + 1} &= -\frac{8}{9} \\ \text{Test} \\ \therefore 7^x - 7^2 = 0 &\text{ or } 3^x - 3^3 = 0 \\ \end{align*}, \begin{align*} The latter gives \(\log(x+1) = 1\), or \(x+1 = 10^{1}\), which admits \(x = 9\). One method is fairly simple but requires a very special form of the exponential equation. 3^{2m} - 27 = 0 \text{ or } & 3^{2m} - 1 = 0\\ All Siyavula textbook content made available on this site is released under the terms of a Because of that all our knowledge about solving equations won’t do us any good. \text{2} & = z -4 \\ 5^{x} - 5 = 0 \text{ or } & 5^{x} + 10 = 0\\ 2^{t} & = 2^{7} \\ 3^{9x - 2} & = 27 \\ In order to determine test values for \(r\) without resorting to the calculator, we need to find numbers between \(0\), \(\frac{1}{e}\), and \(1\) which have a base of \(e\). We now have the same base and a single exponent on each base so we can use the property and set the exponents equal. y^{\frac{1}{2}} - 1 & = 0\\ Doing this gives. Remember, an exponential equation has the variable in the exponent. Let’s start off by looking at the simpler method. This reduces to \(x^2+x-6 = 0\), which gives \(x=-3\) and \(x=2\). However, we could have arrived at the same answer, in fewer steps, by using Theorem \ref{invpropslogs} to rewrite the equation \(\log_{2}(x) = 3\) as \(2^{3} = x\), or \(x=8\). 4^{x + 3} & = \text{0,5} \\ The \(x\) in now out of the exponent! 2^{t} & = 2^{3}\\ 27^{x} \times 9^{x - 2} & = 1 \\ Therefore: Join thousands of learners improving their maths marks online with Siyavula Practice. 2^3/2 is √(2³), or √8 = 2.83. y & = -7 In this part we’ve got some issues with both sides. Once this is done we then factor out a \(y\) and divide by the coefficient. An exponential equation is an equation in which the variable is in the exponent. We define \(r(x) = x \log(x+1) - x \( and due to the presence of the logarithm, we require \(x+1 > 0\), or \(x > -1\). Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. Before we can combine the logarithms, however, we need a common base. From our sign diagram, we find the solution to be \(\left(0, \frac{1}{e}\right) \cup [1, \infty)\). So, we get all the terms with \(y\) in them on one side and all the other terms on the other side. \therefore 9x - 2 & = 3 \\ Now, this one is a little easier than the previous one. This is a fairly simple equation however the method we used in the previous examples just won’t work because we don’t know how to write 9 as a power of 7. Solve for \(x\): \[3^{2x}-80\cdot 3^{x} - 81 = 0\]. \end{align*}, \(- \frac{1}{2} \cdot 6^{\frac{m}{2} + 3} = -18\), \begin{align*} 2^{a} & = 2^{-3}\\ \frac{2^{3x} - 1}{2^x - 1} &= 8.2^x + 9 \\ Doing this gives. 2^{x + 5} & = 32 \\ Let’s take a look at a couple of examples. \therefore x & = -3 \(3^x = 81\) or \(3^x = -1\). \item Moving all of the nonzero terms of \(\left(\log_{2}(x)\right)^2 < 2 \log_{2}(x) + 3\) to one side of the inequality, we have \(\left(\log_{2}(x)\right)^2 - 2 \log_{2}(x) - 3 < 0\). 6^{\frac{m}{2} + 3} & = 36 \\ \left(y^{\frac{1}{2}} - 1\right)\left(y^{\frac{1}{2}} - 1\right) & = 0 \\ Now, in this case we don’t have the same base so we can’t just set exponents equal. \therefore 10^x - 10^0 = 0 &\text{ or } 3^x - 3^4 = 0 \\ Okay, so we say above that if we had a logarithm in front the left side we could get the \(x\) out of the exponent. 2^{t}(1 + 2^{2}) & = 40\\ \end{align*}, \begin{align*} This is one method to solve exponential equations. \end{align*}. You may choose to return to this example once the \(k\)-substitution has been taught. x & = -\dfrac{5}{2} (x - 3)(x + 1) & = 0\\ We use test values which are powers of \(2\): \(0 < \frac{1}{4} < \frac{1}{2} < 1 < 8 < 16\), and from our sign diagram, we see \(r(x)< 0\) on \(\left(\frac{1}{2}, 8 \right)\). We can use either logarithm, although there are times when it is more convenient to use one over the other. We can only use the facts to simplify this if there isn’t a coefficient on the exponential. 5^{2 - 4x} & = 5^{4} \\ Taking the intersection gives us our final answer \(10^{-8.5} \leq [\mbox{H}^{+}] \leq 10^{-7.8}\). (10^x - 1)(3^x - 81) &= 0 \\ Here’s what we get when we use this fact. \end{align*}, \begin{align*} \frac{(2^{x} - 1)(2^{2x} + 2^x + 1)}{2^x - 1} &= 8.2^x + 9 \\ We set \(u = \log_{2}(x)\) so our equation becomes \(u^2-2u-3 = 0\) which gives us \(u=-1\) and \(u=3\). Creative Commons Attribution License. We then use the Quotient Rule and convert to an exponential equation \[\log_{2}\left(\frac{x+3}{6-x}\right) = 3 \iff 2^{3} = \frac{x+3}{6-x} \] This reduces to the linear equation \(8(6-x) = x+3\), which gives us \(x = 5\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. With this final equation we’ve got a couple of issues. 10^{x} & = \text{0,001} \\ Also, be careful here to not make the following mistake. That is perfectly acceptable so don’t worry about it when it happens. 2^{\text{2,805}} &= \text{6,989} \\ To solve an exponential equation, take the log of both sides, and solve for the variable. Next, in order to move the exponent down it has to be on the whole term inside the logarithm and that just won’t be the case with this equation in its present form.