Logaritme- og eksponentiallikninger

$$ $ \begin{array}{rcl}\lg x & =& 5\\ x & >& 0\\ \lg x & =& 5\\ x & =& {10}^5 =100 000\\ & & \end{array}$$$

Løst med CAS:

$$ $ $ \begin{array}{rcl}\lg 4x-\lg x& =& x\\ \lg 4+\lg x-\lg x& =& x\\ x& = & \lg 4\\ x& \approx & 0,60\end{array}$$$$

Løst med CAS:

$$ $ \begin{array}{rcl}3\lg x-\lg x-1 & =& 1\\ x & >& 0\\ 3\lg x-\lg x-1 & =& 1\\ 2\lg x & =& 2\\ \lg x & =& 1\\ x & =& {10}^1 = 10\\ & & \end{array}$$$

Løst med CAS:

$$ $ \begin{array}{rcl}\lg (x+2) & =& 4\\ x & >& -2\\ \lg (x+2) & =& 4\\ x+2 & =& {10}^4\\ x & =& 9998\\ & & \end{array}$$$

$$ $ \begin{array}{rcl}\lg x^2-\lg x & =& 2\\ x & >& 0\\ 2\lg x-\lg x & =& 2\\ \lg x & =& 2\\ x & =& {10}^2\\ x & =& 100\\ & & \end{array}$$$

$$ $ \begin{array}{rcl}\lg (x^2-1)-\lg (x-1) & =& 2\\ x & >& 1\\ \lg \frac{x^2+1}{x-1} & =& 2\\ \lg \left(\frac{(x+1)\cancel{(x-1)}}{\cancel{x-1}}\right) & =& 2\\ \lg (x+1) & =& 2\\ {10}^{\lg (x+1)} & =& {10}^2\\ x+1 & =& 100\\ x & =& 99\\ & & \end{array}$$$

$$ $ \begin{array}{rcl}\lg \left(\frac{x}{2}\right)+\lg \left(\frac{x}{8}\right) & =& 0\\ \lg x-\lg 2+\lg x-\lg 8 & =& 0\\ 2\lg x-\lg 2-\lg 2^3 & =& 0\\ 2\lg x -\lg 2-3\lg 2& =& 0\\ 2\lg x & =& 4\lg 2\\ \lg x & =& 2\lg 2\\ \lg x & =& \lg 2^2\\ \lg x & =& \lg 4\\ x & =& 4\end{array}$$$

$$ $ \begin{array}{rcl}\lg (3x+1)-\lg (x+5) & =& 0\\ \lg (3x+1) & =& \lg (x+5)\\ 3x+1 & =& x+5\\ 2x & =& 4\\ x & =& 2\\ & & \end{array}$$$

$$ \lg (x+3)$$ er gyldig for $$ x>-3$$.

$$ \begin{array}{rcl}\lg (x+3) & =& 1\\ {10}^{\lg (x+3)} & =& {10}^1\\ x+3 & =& 10\\ x & =& 7\end{array}$$

Løsning med CAS:

$$ \lg \left(2x-1\right) \mathrm{er} \mathrm{gyldig} \mathrm{for} x>\frac{1}{2}.$$

$$ \begin{array}{rcl}\lg (2x-1) & =& 2\\ {10}^{\lg (2x-1)} & =& {10}^2\\ 2x-1 & =& {10}^2\\ x & =& \frac{100+1}{2}\\ x & =& 50,5\end{array}$$

Løsning med CAS:

$$ \lg x \mathrm{er} \mathrm{gyldig} \mathrm{for} x>0.$$

$$ \begin{array}{rcl}\lg x & =& \frac{-5\pm \sqrt{5^2-4·1·(-6)}}{2·1}\\ \lg x & =& \frac{-5\pm 7}{2}\\ \lg x & =& 1 \vee \lg x=-6\\ x & =& 10 \vee x=0,000 001\end{array}$$

Løsning med CAS (merk at vi bare får med én av løsningene):

$$ \lg x \mathrm{er} \mathrm{gyldig} \mathrm{for} x>0.$$

$$ \begin{array}{rcl}2\lg x+5\lg x-6 & =& 0\\ 7\lg x & =& 6\\ \lg x & =& \frac{6}{7}\\ x & =& {10}^{\frac{6}{7}}\approx 7,2\end{array}$$

Løsning med CAS:

$$ \lg x \mathrm{er} \mathrm{gyldig} \mathrm{for} x>0.$$

$$ \begin{array}{rcl}\lg x^4-\lg x & =& 18\\ 4\lg x-\lg x & =& 18\\ 3\lg x & =& 18\\ \lg x & =& 6\\ {10}^{\lg x} & =& {10}^6\\ x & =& 1 000 000\end{array}$$

Løsning med CAS:

$$\begin{gathered} \ln \left(4x-2\right) \mathrm{er} \mathrm{gyldig} \mathrm{for} x>\frac{1}{2} \\ \end{gathered}$$, og

$$ \ln (2x-2) \mathrm{er} \mathrm{gyldig} \mathrm{for} x>1$$.

Det betyr at vi kun har løsning når $$ x>1.$$

$$ \begin{array}{rcl}\ln (4x-\mathrm{2)}-\ln (2x-2)-2& = & 0\\ \ln \frac{4x-2}{2x-2}& =& 2\\ e^{\ln \frac{2x-1}{x-1}}& =& e^2\\ \frac{2x-1}{x-1}& =& e^2\\ 2x-1& =& e^2·(x-1)\\ 2x-e^{2}x& =& 1-e^2\\ x(2-e^2)& =& 1-e^2\\ x& =& \frac{1-e^2}{(2-e^2)}\approx 1,19\end{array}$$

Løsning med CAS:

$$\begin{gathered} \lg (x+2) \mathrm{er} \mathrm{gyldig} \mathrm{for} x>-2 \\ \end{gathered}$$, og

$$ \lg (x-1) \mathrm{er} \mathrm{gyldig} \mathrm{for} x>1$$.

Det betyr at vi kun har løsning når $$ x>1.$$

$$ \begin{array}{rcl}\lg (x+2)+\lg (x-1) & =& 1\\ \lg \left((x+2)·(x-1)\right) & =& 1\\ {10}^{\lg \left((x+2)·(x-1)\right)} & =& {10}^1\\ (x+2)(x-1) & =& 10\\ x^2-x+2x-2 & =& 10\\ x^2+x-12 & =& 0\\ (x-3)(x+4) & =& 0\\ x & =& 3 \vee x=-4\end{array}$$

$$ x=-4$$ går ikke, siden det gir logaritmen til et negativt tall.

Løsningen er derfor $$ x=3.$$

Løsning med CAS:

$$\begin{gathered} \ln \left(3x+2\right) \mathrm{er} \mathrm{gyldig} \mathrm{for} x>-\frac{2}{3} \\ \end{gathered}$$, og

$$ \ln 2x \mathrm{er} \mathrm{gyldig} \mathrm{for} x>0$$.

Det betyr at vi kun har løsning når $$ x>0.$$

$$ \begin{array}{rcl}\ln \frac{3x+2}{2x} & =& 0\\ \frac{3x+2}{2x} & =& 1\\ 3x+2 & =& 2x\\ x & =& -2\end{array}$$

Vi har at $$ x<0$$, dermed har likningen ingen løsning.

Løsning med CAS:

$$ \begin{array}{rcl}{10}^x& = & 100\\ \lg {10}^x& =& \lg 100\\ x\lg 10& =& \lg 100\\ x& =& \frac{\lg 100}{\lg 10}\\ x& =& \frac{2}{1}\\ x& =& 2\end{array}$$

$$ \begin{array}{rcl}{10}^{-2x} & =& 100\\ -2x\lg 10 & =& 2\lg 10\\ -2x & =& 2\\ x & =& -1\end{array}$$

$$ \begin{array}{rcl}{5}^x & =& 125\\ \ln 5 & =& \ln 125\\ x\ln 5 & =& 3\ln 5\\ x & =& \frac{3\ln 5}{\ln 5}\\ x & =& 3\end{array}$$

$$ \begin{array}{rcl}\mathrm{2,0}·0,5^x & =& 16\\ \lg (\mathrm{2,0}·0,5^x) & =& \lg 16\\ \lg 2+\lg 0,5^x & =& \lg 16\\ \lg 2+x\lg 0,5 & =& \lg 16\\ \lg 2+x\lg \frac{1 }{2}& =& \lg 2^4\\ \lg 2+x(\lg 1-\lg 2) & =& 4\lg 2\\ x(\lg 1-\lg 2) & =& 4\lg 2-\lg 2\\ x & =& \frac{3\lg 2}{\lg 1-\lg 2}\\ x & =& \frac{3\cancel{\lg 2}}{-\cancel{\lg 2}}=-3\end{array}$$

$$ \begin{array}{rcl}\frac{1}{4}·4^{2x}& = & 16\\ 4^{2x} & =& 4^2·4\\ 4^{2x}& =& 4^3\\ 2x& =& 3\\ x& =& \frac{3}{2}\end{array}$$

$$ \begin{array}{rcl}\cancel{4}·6^x& =& \cancel{36}·2^x\\ 6^x& = & 9·2^x\\ (2·3{)}^x& = & 3^2·2^x\\ \cancel{2^x}·3^x & = & 3^2·\cancel{2^x}\\ 3^{x }& = & 3^2\\ x& = & 2\end{array}$$

$$ \begin{array}{rcl}{2}^{2x}+2^x-6& = & 0\\ {\left(2^x\right)}^2+2^x-6& =& 0\end{array}$$

Vi bruker abc-formelen for andregradslikninger.

$$ \begin{array}{rcl}(2^x{)}^2+2^x-6& = & 0\\ 2^x& =& \frac{-1\pm \sqrt{1^2-4·1·(-6)}}{2·1}\\ 2^x& =& \frac{-1\pm \sqrt{25}}{2}\\ 2^x& =& \frac{-1+5}{2} \vee 2^x = \frac{-1-5}{2}\\ 2^x& =& 2 \vee 2^x = -3\end{array}$$

$$ 2^x=-3 \mathrm{har} \mathrm{ingen} \mathrm{løsning}, \mathrm{siden} 2^x>0 \mathrm{for} \mathrm{alle} x.$$

$$ \begin{array}{rcl}{2}^x & =& 2\\ x & =& 1\end{array}$$

$$ \begin{array}{rcl}{e}^x-6e^{-x} & =& 1\\ e^x·e^x-\frac{6}{e^x}·e^{x }& =& 1·e^x\\ (e^x{)}^2-6 & =& e^x\\ (e^x{)}^2-e^x-6 & =& 0\\ e^x & =& \frac{1\pm \sqrt{(-1{)}^2-4·1·(-6)}}{2·1}\\ e^x & =& \frac{1\pm \sqrt{25}}{2}\\ e^x & =& 3 \vee e^x=-2 \\ & & (\mathrm{ingen} \mathrm{løsning} \mathrm{for} e^x \mathrm{mindre} \mathrm{enn} 0)\end{array}$$

$$ \begin{array}{rcl}{4}^{2x}-2·4^x-3 & =& 0\\ (4^x{)}^2-2·4^x-3 & =& 0\\ 4^x & =& \frac{2\pm \sqrt{(-2{)}^2-4·1·(-3)}}{2·1}\\ 4^x & =& \frac{2\pm \sqrt{4+12}}{2}\\ 4^x & =& 3 \vee \cancel{4^x=-1}\\ \lg 4^x & =& \lg 3\\ x & =& \frac{\lg 3}{\lg 4}\end{array}$$

$$ \begin{array}{rcl}\lg x & =& 2,24\\ {10}^{\lg x} & =& {10}^{2,24}\\ x & =& 173,78\end{array}$$

$$ \begin{array}{rcl}\ln x & =& -1,85\\ e^{\ln x} & =& e^{-1,85}\\ x & =& 0,16\end{array}$$

$$ \begin{array}{rcl}2·\lg x & =& 0,24\\ \lg x & =& \frac{0,24}{2}\\ {10}^{\lg x} & =& {10}^{0,12}\\ x & =& 1,32\end{array}$$

$$ \begin{array}{rcl}2·\lg x+0,12 & =& 0,24\\ \lg x+0,06 & =& 0,12\\ \lg x & =& 0,06\\ {10}^{\lg x }& =& {10}^{0,06}\\ x & =& 1,15\end{array}$$

$$ \begin{array}{rcl}2·\ln x-2,0 & =& 0\\ \ln x-1 & =& 0\\ \ln x & =& 1\\ e^{\ln x} & =& e^1\\ x & =& e\end{array}$$

$$ \begin{array}{rcl}2·\lg x^2-3\lg x & =& 0,24\\ 2·2\lg x-3\lg x & =& 0,24\\ \lg x & =& 0,24\\ {10}^{\lg x }& =& {10}^{0,24}\\ x & =& 1,74\end{array}$$

$$ \begin{array}{rcl}\lg x^2-\lg 2x & =& \lg 8\\ 2\lg x-(\lg 2+\lg x) & =& \lg 8\\ 2\lg x-\lg 2-\lg x & =& \lg 2^3\\ \lg x & =& 3\lg 2+\lg 2\\ & =& 4\lg 2\\ & =& \lg 2^4\\ {10}^{\lg x} & =& {10}^{\lg 2^4}\\ x & =& 16\end{array}$$

$$ \begin{array}{rcl}\lg \frac{x}{2}-2\lg x+\lg 4 & =& 0\\ \lg x-\lg 2-2\lg x+\lg 2^2 & =& 0\\ -\lg x-\lg 2+2\lg 2 & =& 0\\ -\lg x+\lg 2 & =& 0\\ \lg 2 & =& \lg x\\ {10}^{\lg 2} & =& {10}^{\lg x}\\ 2 & =& x\end{array}$$

Løsning på oppgave 1:

$$ \begin{array}{rcl}{7}^x & =& 3^{x+1}\\ \lg 7^x & =& \lg 3^{x+1}\\ x\lg 7 & =& (x+1)\lg 3\\ x\lg 7 & =& x\lg 3+\lg 3\\ x\lg 7-x\lg 3 & =& \lg 3\\ x(\lg 7-\lg 3) & =& \lg 3\\ x & =& \frac{\lg 3}{\lg 7-\lg 3}\end{array}$$

Løsning på oppgave 2: Se Idas løsningsforslag.

Løsning på oppgave 3: Se Sanders løsningsforslag.